Does it have exactly $n$ roots? Would replacing the R.H.S. of the equation with any other real number change the outcome? I can show that the equation has no complex roots. But how to find the number of roots?
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Let $f(x)$ be the left side. Note that $f(x) \to 0$ as $x \to \pm \infty$, $+\infty$ as $x \to a_i-$ and $-\infty$ as $x \to a_i+$. So there will be at least one root in each interval $(-\infty, a_1)$, $(a_1, a_2)$, ..., $(a_{n-1}, a_n)$. Since the equation is equivalent to a polynomial equation with degree $n$, there are at most $n$ roots. Therefore there are exactly $n$ real roots.
EDIT: There is nothing special about $2016$: any other positive real number would work (for a negative number, you would replace $(-\infty, a_1)$ by $(a_n, \infty)$ and still have $n$ real roots). For $0$, on the other hand, there would only be $n-1$ roots.
Robert Israel
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What about when the r.h.s. is simply $x$? – StubbornAtom Jul 21 '16 at 19:38
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@StubbornAtom Then you still have a root in $(a_i, a_{i+1})$ for each $i$, and none in $(a_n, \infty)$, but you may or may not have two in $(-\infty, a_1)$. – Robert Israel Jul 21 '16 at 20:47
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Sorry again, but if r.h.s. is $n$ then does it mean that the equation has at least $n-1$ roots? – StubbornAtom Nov 18 '16 at 13:10
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Presumably $n$ is a positive real number, so there are exactly $n$ roots. – Robert Israel Nov 18 '16 at 16:38
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The equation is nothing but a polynomial of degree $n$ in disguise (except at $x=a_i$). Hence, it will have $n$ roots. You can show that the roots are interlaced by consecutive $a_i$'s and hence will have $n$ real roots. We have $$p(x) = 2016 \prod_{k=1}^n (a_k-x) - \sum_{i=1}^n \left(a_i \prod_{\overset{k=1}{k \neq i}}^n (a_k-x) \right)$$ We have $$p(a_1) < 0, p(a_2)>0, p(a_3)<0, \cdots,$$ This ensures there are $n-1$ real roots that are interlaced. Hence, the $n^{th}$ root must also be real.
Adhvaitha
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Your second sentence is somewhat imprecise. The polynomial is of degree $n$, so it has $n$ not necessarily distinct, possibly complex roots (by the fundamental theorem of algebra). – Ewan Delanoy Jul 21 '16 at 17:17