This is a really neat exercise. Here is the answer:
Theorem 1. Let $n\in\mathbb{N}$. (Here, as always, $\mathbb{N}=\left\{
0,1,2,\ldots\right\} $.) Let $m=\left\lfloor \left( n+1\right)
/2\right\rfloor $. Then,
\begin{equation}
\sum\limits_{k=0}^{n}\left( -1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}=m\left(
-1\right) ^{m}\dbinom{x}{m}
\end{equation}
as polynomials in $\mathbb{Q}\left[ x\right] $.
Note that my $x$, $n$ and $k$ are your $n$, $k$ and $i$ (sorry for this -- I
am taking the lazy route and adapting your notations to mine), and I have
extended the domains for $x$ (promoted from a lowly integer to a polynomial
indeterminate) and $n$ (now any nonnegative integer).
The proof will rely on the following two facts:
Lemma 2. Let $k$ be a positive integer. Then,
\begin{equation}
k\dbinom{x}{k}=x\dbinom{x-1}{k-1}\qquad\text{as polynomials in }
\mathbb{Q}\left[ x\right] .
\end{equation}
Proof of Lemma 2. This is usually stated in the equivalent form $\dbinom
{x}{k}=\dfrac{x}{k}\dbinom{x-1}{k-1}$; in this form it is:
You will have likely proven it by the time you have found it in these sources.
Note that this identity is the key to algebraic proofs of various identities
with "$k\dbinom{x}{k}$"s in them -- such as
$\sum\limits_{k=0}^{n}k\dbinom{n}{k}=n2^{n-1}$ and $\sum\limits_{k=0}^{n}\left( -1\right)
^{k}k\dbinom{n}{k}=
\begin{cases}
-1, & \text{if }n=1;\\
0, & \text{if }n\neq1
\end{cases}
$ for all $n\in\mathbb{N}$. $\blacksquare$
Lemma 3. Let $n\in\mathbb{N}$. Then,
\begin{equation}
\sum\limits_{k=0}^{n}\left( -1\right) ^{k}\dbinom{x}{k}\dbinom{x}{n-k}=
\begin{cases}
\left( -1\right) ^{n/2}\dbinom{x}{n/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd}
\end{cases}
\label{darij1.eq.l3.eq}
\tag{1}
\end{equation}
as polynomials in $\mathbb{Q}\left[ x\right] $.
Proof of Lemma 3. This is Exercise 3.22 in my Notes on the combinatorial
fundamentals of algebra, version of 10 January
2019.
Alternatively, if $x$ is specialized to a nonnegative integer, you can use
Mike Spivey's argument at
Alternating sum of squares of binomial coefficients
(which is stated for the particular case $n=x$, but can easily be adapted to
the general case -- see my comment under his post) to prove
\eqref{darij1.eq.l3.eq} combinatorially; then, use the "polynomial identity
trick" to un-specialize $x$. You can probably find lots of other approaches on
math.stackexchange. Either way, Lemma 3 is proven. $\blacksquare$
Now, we can prove Theorem 1:
Proof of Theorem 1. It is easy to prove Theorem 1 in the case when $n=0$.
(Indeed, in this case, both sides of the equality in question equal $0$, since
they are products in which one of the factors is $0$.) Thus, for the rest of
this proof, we WLOG assume that $n\neq0$. Hence, $n>0$. Thus,
$n-1 \in \mathbb{N}$.
We shall use the convention that $\dbinom{u}{v}=0$ whenever $v\notin
\mathbb{N}$. Thus, the recurrence of the binomial coefficients,
\begin{equation}
\dbinom{u}{v}=\dbinom{u-1}{v-1}+\dbinom{u-1}{v},
\label{darij1.pf.t1.1}
\tag{2}
\end{equation}
holds not only for $v\in\left\{ 1,2,3,\ldots\right\} $ but for all
$v\in\mathbb{Z}$.
Lemma 3 (applied to $n-1$ instead of $n$) yields
\begin{align*}
\sum\limits_{k=0}^{n-1}\left( -1\right) ^{k}\dbinom{x}{k}\dbinom{x}{\left(
n-1\right) -k} & =
\begin{cases}
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x}{\left( n-1\right)
/2}, & \text{if }n-1\text{ is even};\\
0, & \text{if }n-1\text{ is odd}
\end{cases}
\\
& =
\begin{cases}
0, & \text{if }n-1\text{ is odd;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x}{\left( n-1\right)
/2}, & \text{if }n-1\text{ is even}
\end{cases}
\\
& =
\begin{cases}
0, & \text{if }n\text{ is even;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd}
\end{cases}
\end{align*}
(since $n-1$ is odd if and only if $n$ is even, and vice versa). Substituting
$x-1$ for $x$ in this equality, we obtain
\begin{equation}
\sum\limits_{k=0}^{n-1}\left( -1\right) ^{k}\dbinom{x-1}{k}\dbinom{x-1}{\left(
n-1\right) -k}=
\begin{cases}
0, & \text{if }n\text{ is even;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x-1}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd.}
\end{cases}
\label{darij1.pf.t1.n-1}
\tag{3}
\end{equation}
If $n>1$, then $n-2\in\mathbb{N}$. Hence, if $n>1$, then Lemma 3 (applied to
$n-2$ instead of $n$) yields
\begin{align*}
\sum\limits_{k=0}^{n-2}\left( -1\right) ^{k}\dbinom{x}{k}\dbinom{x}{\left(
n-2\right) -k} & =
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x}{\left( n-2\right)
/2}, & \text{if }n-2\text{ is even};\\
0, & \text{if }n-2\text{ is odd}
\end{cases}
\\
& =
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x}{\left( n-2\right)
/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd}
\end{cases}
\end{align*}
(since $n-2$ is even if and only if $n$ is even, and since $n-2$ is odd if and
only if $n$ is odd). This equality holds not only for $n>1$, but also for
$n=1$ (since both of its sides equal $0$ in this case), and thus holds in all
cases (since we have $n\geq1$). Substituting $x-1$ for $x$ in this equality,
we obtain
\begin{equation}
\sum\limits_{k=0}^{n-2}\left( -1\right) ^{k}\dbinom{x-1}{k}\dbinom{x-1}{\left(
n-2\right) -k}=
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}
\end{cases}
\end{equation}
The left hand side of this equality does not change if we
replace the summation sign "$\sum\limits_{k=0}^{n-2}$" by "$\sum\limits_{k=0}^{n-1}$"
(because the only new addend that we gain in this way is
$\left( -1\right) ^{n-1}\dbinom{x-1}{n-1}
\underbrace{\dbinom{x-1}{\left( n-2\right) -\left(n-1\right)}}_{\substack{
= 0 \\ \text{(since $\left(n-2\right)-\left(n-1\right) = -1 \notin \mathbb{N}$)}}}
= 0$). Hence, this equality becomes
\begin{equation}
\sum\limits_{k=0}^{n-1}\left( -1\right) ^{k}\dbinom{x-1}{k}\dbinom{x-1}{\left(
n-2\right) -k}=
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd.}
\end{cases}
\label{darij1.pf.t1.n-2}
\tag{4}
\end{equation}
We can split off the addend for $k=0$ from the sum $\sum\limits_{k=0}^{n}\left(
-1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}$ (since $n\geq0$). Thus, we find
\begin{align}
& \sum\limits_{k=0}^{n}\left( -1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}
\nonumber\\
& =\underbrace{\left( -1\right) ^{0}0\dbinom{x}{0}\dbinom{x}{n-0}}_{=0}
+\sum\limits_{k=1}^{n}\left( -1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}
\nonumber\\
& =\sum\limits_{k=1}^{n}\underbrace{\left( -1\right) ^{k}}_{=-\left( -1\right)
^{k-1}}\underbrace{k\dbinom{x}{k}}_{\substack{=x\dbinom{x-1}{k-1}\\\text{(by
Lemma 2)}}}\underbrace{\dbinom{x}{n-k}}_{\substack{=\dbinom{x-1}
{n-k-1}+\dbinom{x-1}{n-k}\\\text{(by \eqref{darij1.pf.t1.1}, applied} \\ \text{to $u = x$ and $v = n-k$)}}}\nonumber\\
& =\sum\limits_{k=1}^{n}\left( -\left( -1\right) ^{k-1}\right) x\dbinom{x-1}
{k-1}\left( \dbinom{x-1}{n-k-1}+\dbinom{x-1}{n-k}\right) \nonumber\\
& =-x\sum\limits_{k=1}^{n}\left( -1\right) ^{k-1}\dbinom{x-1}{k-1}\left(
\dbinom{x-1}{n-k-1}+\dbinom{x-1}{n-k}\right) .
\label{darij1.pf.t1.4}
\tag{5}
\end{align}
Now,
\begin{align*}
& \sum\limits_{k=1}^{n}\left( -1\right) ^{k-1}\dbinom{x-1}{k-1}\left( \dbinom
{x-1}{n-k-1}+\dbinom{x-1}{n-k}\right) \\
& =\sum\limits_{k=0}^{n-1}\left( -1\right) ^{k}\dbinom{x-1}{k}\left(
\underbrace{\dbinom{x-1}{n-k-2}}_{=\dbinom{x-1}{\left( n-2\right) -k}
}+\underbrace{\dbinom{x-1}{n-k-1}}_{=\dbinom{x-1}{\left( n-1\right) -k}
}\right) \\
& \qquad\left( \text{here, we have substituted }k+1\text{ for }k\text{ in the
sum}\right) \\
& =\underbrace{\sum\limits_{k=0}^{n-1}\left( -1\right) ^{k}\dbinom{x-1}{k}
\dbinom{x-1}{\left( n-2\right) -k}}_{\substack{=
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd}
\end{cases}
\\\text{(by \eqref{darij1.pf.t1.n-2})}}}+\underbrace{\sum\limits_{k=0}^{n-1}\left(
-1\right) ^{k}\dbinom{x-1}{k}\dbinom{x-1}{\left( n-1\right) -k}
}_{\substack{=
\begin{cases}
0, & \text{if }n\text{ is even;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x-1}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd}
\end{cases}
\\\text{(by \eqref{darij1.pf.t1.n-1})}}}\\
& =
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}, & \text{if }n\text{ is even};\\
0, & \text{if }n\text{ is odd}
\end{cases}
+
\begin{cases}
0, & \text{if }n\text{ is even;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x-1}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd}
\end{cases}
\\
& =
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}+0, & \text{if }n\text{ is even;}\\
0+\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x-1}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd}
\end{cases}
\\
& =
\begin{cases}
\left( -1\right) ^{\left( n-2\right) /2}\dbinom{x-1}{\left( n-2\right)
/2}, & \text{if }n\text{ is even;}\\
\left( -1\right) ^{\left( n-1\right) /2}\dbinom{x-1}{\left( n-1\right)
/2}, & \text{if }n\text{ is odd}
\end{cases}
\\
& =
\begin{cases}
\left( -1\right) ^{\left\lfloor \left( n-1\right) /2\right\rfloor }
\dbinom{x-1}{\left\lfloor \left( n-1\right) /2\right\rfloor }, & \text{if
}n\text{ is even;}\\
\left( -1\right) ^{\left\lfloor \left( n-1\right) /2\right\rfloor }
\dbinom{x-1}{\left\lfloor \left( n-1\right) /2\right\rfloor }, & \text{if
}n\text{ is odd}
\end{cases}
\\
& \qquad\left(
\begin{array}
[c]{c}
\text{since }\left( n-2\right) /2=\left\lfloor \left( n-1\right)
/2\right\rfloor \text{ when }n\text{ is even,}\\
\text{and since }\left( n-1\right) /2=\left\lfloor \left( n-1\right)
/2\right\rfloor \text{ when }n\text{ is odd}
\end{array}
\right) \\
& =\left( -1\right) ^{\left\lfloor \left( n-1\right) /2\right\rfloor
}\dbinom{x-1}{\left\lfloor \left( n-1\right) /2\right\rfloor }.
\end{align*}
Thus, \eqref{darij1.pf.t1.4} becomes
\begin{align}
& \sum\limits_{k=0}^{n}\left( -1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}
\nonumber\\
& =-x\underbrace{\sum\limits_{k=1}^{n}\left( -1\right) ^{k-1}\dbinom{x-1}
{k-1}\left( \dbinom{x-1}{n-k-1}+\dbinom{x-1}{n-k}\right) }_{=\left(
-1\right) ^{\left\lfloor \left( n-1\right) /2\right\rfloor }\dbinom
{x-1}{\left\lfloor \left( n-1\right) /2\right\rfloor }}\nonumber\\
& =-x\left( -1\right) ^{\left\lfloor \left( n-1\right) /2\right\rfloor
}\dbinom{x-1}{\left\lfloor \left( n-1\right) /2\right\rfloor } .
\label{darij1.pf.t1.7}
\tag{6}
\end{align}
On the other hand, recall that $m=\left\lfloor \left( n+1\right)
/2\right\rfloor $, so that $m-1=\left\lfloor \left( n+1\right)
/2\right\rfloor -1=\left\lfloor \underbrace{\left( n+1\right) /2-1}
_{=\left( n-1\right) /2}\right\rfloor =\left\lfloor \left( n-1\right)
/2\right\rfloor $. Also, $m=\left\lfloor \left( n+1\right) /2\right\rfloor
\geq1$ (since $n\geq1$ and thus $\left( n+1\right) /2\geq1$). Hence, $m$ is
a positive integer; thus, Lemma 2 (applied to $k=m$) yields $m\dbinom{x}
{m}=x\dbinom{x-1}{m-1}$. Now,
\begin{align*}
m\left( -1\right) ^{m}\dbinom{x}{m} & =\underbrace{\left( -1\right) ^{m}
}_{=-\left( -1\right) ^{m-1}}\underbrace{m\dbinom{x}{m}}_{=x\dbinom
{x-1}{m-1}}=-\left( -1\right) ^{m-1}x\dbinom{x-1}{m-1}\\
& =-x\left( -1\right) ^{m-1}\dbinom{x-1}{m-1}=-x\left( -1\right)
^{\left\lfloor \left( n-1\right) /2\right\rfloor }\dbinom{x-1}{\left\lfloor
\left( n-1\right) /2\right\rfloor }
\end{align*}
(since $m-1=\left\lfloor \left( n-1\right) /2\right\rfloor $). Comparing
this with \eqref{darij1.pf.t1.7}, we obtain
\begin{equation}
\sum\limits_{k=0}^{n}\left( -1\right) ^{k}k\dbinom{x}{k}\dbinom{x}{n-k}=m\left(
-1\right) ^{m}\dbinom{x}{m}.
\end{equation}
This proves Theorem 1. $\blacksquare$
