The expression under the limit sign equals (exactly)
$$f_N(\delta):=\sum_{n=1}^{2N}(-1)^{n-1}\sqrt{\frac{\delta^2}{4}+N(N+\delta)\sin^2\frac{n\pi}{2N}}.$$
As $\sum_{n=1}^{2N}(-1)^{n-1}\sin\frac{n\pi}{2N}=\tan\frac{\pi}{4N}$ (yes, the case $\delta=0$ is easy), we can write
$$f_N(\delta)=\Big(N+\frac{\delta}{2}\Big)\tan\frac{\pi}{4N}-\frac{|\delta|}{2}+2\sum_{n=1}^{N-1}(-1)^{n-1}\times{}\\{}\times\left(\sqrt{\frac{\delta^2}{4}+N(N+\delta)\sin^2\frac{n\pi}{2N}}-\Big(N+\frac{\delta}{2}\Big)\sin\frac{n\pi}{2N}\right).$$
Here, the limit can be taken termwise — after pairing "odds and evens", this is justified by (a discrete version of) the dominated convergence theorem. This immediately gives
$$\color{blue}{f(\delta)=\frac{\pi}{4}-\frac{|\delta|}{2}+\sum_{n=1}^{\infty}(-1)^{n-1}\big(\sqrt{\delta^2+n^2\pi^2}-n\pi\big)}$$
which also has an integral representation: $\color{blue}{f(\delta)=\displaystyle\frac{1}{\pi}\int_{|\delta|}^{\infty}\frac{\sqrt{x^2-\delta^2}}{\sinh x}\,dx}$ (see the linked question). Below is my older answer obtaining this integral another (somewhat more complicated) way.
$$f_N(\delta)=\frac{N+\delta}{2}\left(g_{2N}\Big(\frac{N}{N+\delta}\Big)-2g_N\Big(\frac{N}{N+\delta}\Big)\right),\qquad(\delta>0)\\ f_N(\delta)=\frac{N}{2}\left(g_{2N}\Big(\frac{N+\delta}{N}\Big)-2g_{N}\Big(\frac{N+\delta}{N}\Big)\right),\qquad(\delta<0)\\g_N(k):=\sum_{n=0}^{N-1}\sqrt{1-2k\cos\frac{2n\pi}{N}+k^2}\qquad(|k|<1)$$
Now the idea is to use Fourier expansion of $\sqrt{1-2k\cos x+k^2}$, together with
$$\sum_{n=0}^{N-1}\cos\frac{2mn\pi}{N}=\begin{cases}N,& N\mid n\\0,& N\not\mid n\end{cases}.$$
The expansion can be obtained indirectly from the binomial series
$$(1-z)^{1/2}=\sum_{m=0}^{\infty}b_m z^m,\qquad b_m=(-1)^m\binom{1/2}{m}=-\frac{1}{\pi}\mathrm{B}\Big(m-\frac{1}{2},\frac{3}{2}\Big)$$
(where $\mathrm{B}$ denotes the beta function; we'll need this later):
\begin{align}
\sqrt{1-2k\cos x+k^2}&=(1-ke^{ix})^{1/2}(1-ke^{-ix})^{1/2}
\\&=\left(\sum_{m=0}^{\infty}b_m k^m e^{imx}\right)\left(\sum_{m=0}^{\infty}b_m k^m e^{-imx}\right)\\&=a_0(k)+2\sum_{m=1}^{\infty}a_m(k)\cos mx,
\\a_m(k)&:=\sum_{n=0}^{\infty}b_n b_{m+n}k^{m+2n}.
\end{align}
So, for $m>0$, using the integral representation of $\mathrm{B}$, we have
\begin{align}
a_m(k)&=-\frac{1}{\pi}\sum_{n=0}^{\infty}b_n k^{m+2n}\int_0^1 t^{m+n-3/2}(1-t)^{1/2}\,dt\\&=-\frac{k^m}{\pi}\int_0^1 t^{m-3/2}\sqrt{(1-t)(1-k^2 t)}\,dt,
\end{align}
which gives
\begin{align}
g_N(k)&=\sum_{n=0}^{N-1}\left(a_0(k)+2\sum_{m=1}^{\infty}a_m(k)\cos\frac{2mn\pi}{N}\right)=Na_0(k)+2N\sum_{m=1}^{\infty}a_{mN}(k)\\&=Na_0(k)-\frac{2N}{\pi}k^N\int_0^1\frac{t^{N-3/2}}{1-(kt)^N}\sqrt{(1-t)(1-k^2 t)}\,dt.
\end{align}
Note that $a_0(k)$ can also be expressed in terms of elliptic integrals:
$$a_0(k)=\frac{2}{\pi}\big(2\mathrm{E}(k)-(1-k^2)\mathrm{K}(k)\big),$$
but we won't need it, since the preceding equality implies
$$g_{2N}(k)-2g_N(k)=\frac{4N}{\pi}k^N\int_0^1\frac{t^{N-3/2}}{1-(kt)^{2N}}\sqrt{(1-t)(1-k^2 t)}\,dt.$$
This can be plugged back into $f_N(\delta)$. For $\delta>0$, substituting $t=1-x/N$, we get
$$f_N(\delta)=\frac{2}{\pi}\Big(\frac{N}{N+\delta}\Big)^N\int_0^N\frac{(1-x/N)^{N-3/2}}{1-\left(\frac{N-x}{N+\delta}\right)^{2N}}\sqrt{x\left(x+2\delta+\frac{\delta^2}{N}\right)}\,dx,$$
and taking $N\to\infty$ is easy now. Doing the same for $\delta<0$, we get the integral above.
