Here is an approach:
We consider the following sum in the limit $n\rightarrow\infty$ (which we denote by $S$) :
$$
S_n=\frac{1}{\binom{2n}{n}}\sum_{k=0}^n(-1)^k\binom{2n}{n+k}\frac{1}{x^2+\pi^2 k^2}
$$
We split the range of summation at some $\delta$ such that $1<<\delta<<n$ to get $S_n=S_{2,n}+S_{1,n}$ and observe that $k<<n$ in $[0,\delta]$ so we can expand the binomial around small $k/n$ which means that
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\left(1+O(k^2/n^3)\right)\frac{1}{x^2+\pi^2 k^2}
$$
since $|\sum_{k=0}^{\delta}(-1)^k\frac{k^2}{x^2+\pi^2 k^2}|<\sum_{k=0}^{\delta}\frac{k^2}{x^2+\pi^2 k^2}<\sum_{k=0}^{\delta}1=\delta$ we can bound the $O(k^2/n^3)$ term as follows
$$
S_{1,n}=\sum_{k=0}^{\delta}(-1)^k\frac{1}{x^2+\pi^2 k^2}+o(\delta/n^3)
$$
Taking limits we get that $S_{1,n}$ approaches a constant to determined later
$$
S_1=\sum_{k=0}^{\infty}(-1)^k\frac{1}{x^2+\pi^2 k^2}\,\,\,\, (*)
$$
Next we show that the tail sum $S_{2,n}$ approaches zero so that the limit in question is indeed given by $(*)$. To this end, observe that on $[\delta,n]$ we have $\pi k >> x$ so we can expand the fraction around large $k$. We find
$$
S_{2,n}=\frac{1}{\binom{2n}{n}}\sum_{k=\delta}^n(-1)^k\binom{2n}{n+k}\left(\frac{1}{\pi^2 k^2}+O(1/k^4)\right)
$$
since the binomial coefficient has a maximum at $k=0$ we can bound $\frac{\binom{2n}{n+k}}{\binom{2n}{n}}\leq1$ so
$$
|S_{2,n}|<\frac{1}{\pi^2}\sum_{k=\delta}^n\frac{1}{k^2}\sim \frac{1}{\pi^2} \int_{\delta}^n\frac{dk}{k^2}\sim \frac{1}{\pi^2\delta}
$$
so, choosing $\delta$ to be a (slowly enough) increasing sequence of $n$ we have indeed that
$$
S_2=0\,\,\,\,(**)
$$
Note that the $O(1/k^4)$ are vanishing even faster and can therefore also be neglected.
Putting anything $(*)$ and $(**)$ together we get
$$
S=S_1=\frac{1}{2x^2}+\frac{1}{2\sinh(x)x}
$$
were the last equality follows from Mittag-Lefflers Theorem (or the product expansion of sine)
Generally similar sums can be evaluated using the Beta function:
$$
B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}=
\frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}.
$$
Applying this in your case ($x=n-k,y=n+k$) one has:
$$
\binom{2n}{n-k}^{-1}=(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt
$$
or
$$\begin{align}
\sum_{n=k}^\infty\binom{2n}{n-k}^{-1}
&=\sum_{n=k}^\infty(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt\\
&=\int_0^1 dt \sum_{n=k}^\infty(2n+1)t^{n-k}(1-t)^{n+k}\\
&=\int_0^1 \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}dt\tag1
\end{align}$$
It can be shown that the integral $(1)$ is a sum of a rational number and a multiple of $\dfrac\pi{9\sqrt3}$.
Indeed:
$$
\frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}=Q_k(t)+\frac{A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3}{(1-t+t^2)^2},\tag2
$$
where both coefficients of the polynomial $Q(t)$ and $A^0,A^1,A^2,A^3$ are integer numbers. The integral of $Q(t)$ is obviously a rational number and
$I_r= \int_0^1\frac{t^r\,dt}{(1-t+t^2)^2}$ can be evaluated as:
$$
I_0=\frac23+\frac49\dfrac\pi{\sqrt3};\quad
I_1=\frac13+\frac29\dfrac\pi{\sqrt3};\quad
I_2=-\frac13+\frac49\dfrac\pi{\sqrt3};\quad
I_3=-\frac23+\frac59\dfrac\pi{\sqrt3}.\quad
$$
Thus the irrational term can be written as:
$$
C_k\frac\pi{9\sqrt3}\quad\text{with}\quad C_k=4A_k^0+2A_k^1+4A_k^2+5A_k^3.
$$
Moreover the term can be evaluated explicitly using the following table:
$$
\begin{array}{c|c|c|c|c|c}
k\mod 3& A_k^0&A_k^1&A_k^2&A_k^3&C_k\\
\hline
0&+1-2x&+1+6x&-1-6x&+0+4x&2\\
1&+2+4x&-5-6x&+3+6x&-1-2x&\hphantom{-1}5+18x\\
2&-3-2x&2&0&-1-2x&-13-18x\\
\hline
\end{array},\tag3
$$
with $x=\left\lfloor\dfrac k3\right\rfloor$, so that $C_k=2,5,-13,2,23,-31,2,41,-49,2,\dots$ for $k=0,1,2,3,4,5,6,7,8,9,\dots$.
The expression $(3)$ can be proved in the following way:
Let
$$
P_k(t)=(1-t)^{2k}[2+(2k-1)(1-t+t^2)];\quad R_k(t)=A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3.\\
$$
Then we have from (2):
$$R_k(t_\pm)=P_k(t_\pm);\quad R'_k(t_\pm)=P'_k(t_\pm),\tag4$$
where
$$
t_\pm=e^{\pm\frac{i\pi}3}
$$
are the roots of the polynomial $t^2-t+1$.
Explicitly (4) amounts to the system of four linear equations:
$$\begin{align}
A_k^0+A_k^1t_\pm+A_k^2t_\pm^2+A_k^3t_\pm^3&=2t_\pm^{-2k}\\
A_k^1+2A_k^2t_\pm+3A_k^3t_\pm^2&=(1-2k-2t_\pm)t_\pm^{-2k}\\
\end{align},
$$
which solutions are given by (3).
Best Answer
Let $\omega$ be a primitive $d$ root of unity. Then define
$$\rm a:=\frac{1}{d}\sum_{j=0}^{d-1} \omega^{\,jk}=\begin{cases}1 & \rm k\equiv 0 ~\bmod d \\ 0 & \rm otherwise \end{cases}=\delta_k$$
To see the equality, first check the $\rm k\equiv0$ case, then the $\rm k\equiv 1$ case by symmetry ($\rm a=\omega \cdot a$), then notice the map $\rm j\mapsto \omega^{jk}$ sends the residues modulo $\rm d$ to the $\rm d/gcd(d,k)$th roots of unity, and it is specifically a $\rm \gcd(d,k)$-to-$1$ map, for any nonzero $\rm k$ (hence $\rm a=0$, going mod $\rm d/(d,k)$). (Also see the Kronecker delta function for a definition of the RHS.)
Using the binomial theorem again with interchange of summation, we have
$$\rm \begin{array}{c c}\frac{1}{d}\sum_{j=0}^{d-1} \big(1+\omega^j\big)^{nd} & \rm =\frac{1}{d}\sum_{j=0}^{d-1}\sum_{k=0}^{nd}\binom{nd}{k} \omega^{\,jk} \\ & \rm =\sum_{k=0}^{nd}\binom{nd}{k}\frac{1}{d}\sum_{j=0}^{d-1}\omega^{\,jk} \\ & \rm =\sum_{k=0}^{nd}\binom{nd}{k}\delta_k \\ & \rm =\sum_{\ell=0}^n \binom{nd}{n\ell}. \end{array}$$
Observe that the first leading term will be with $\omega=1$. Set $\rm \omega=e^{2\pi i/d}$; ordering $|1+\omega^j|$ by magnitude, we find the $(v+1)$th leading term is (for $v\le d/2$)
$$\begin{array}{c c} \rm \frac{(1+\omega^v)^{nd}+(1+\omega^{-v})^{nd}}{d} & = \rm \frac{(\omega^{v/2}+\omega^{-v/2})^{nd}(\omega^{vnd/2}+\omega^{-vnd/2})}{d} \\ & =\rm \frac{2}{d}\big(2\cos(2\pi v/d) \big)^{nd}\cos(\pi vnd)\end{array}$$
Similar considerations lead to an offset modular generalization:
$$\rm \frac{1}{d}\sum_{j=0}^{d-1}\omega^{-j\,r} \big(1+\omega^j\big)^{nd} =\sum_{\ell=0}^n \binom{nd}{n\ell+r}.$$