Let's present a generalization and also give an explicit solution.

If $M$ is the $n\times n$ matrix
$$M=\left[\frac{1+(-1)^{i+j}}{x_i-y_j}\right]_{i,j=1}^n$$
then the inverse matrix $K:=M^{-1}$ has entries given by
\begin{align}
K_{a,b}=\begin{cases}
2\frac{\prod_{2j-1\neq b}x_{2j-1}-y_a}{\prod_{2j-1\neq a}y_a-y_{2j-1}}\cdot \frac{\prod_{2k-1}x_b-y_{2k-1}}{\prod_{2k-1\neq b}x_{2k-1}-x_b} \qquad \text{$a, b$ are odd} \\
\,\,\,\,\,\,\,\, 2\frac{\prod_{2j\neq b}x_{2j}-y_a}{\prod_{2j\neq a}y_a-y_{2j}}\cdot \frac{\prod_{2k}x_b-y_{2k}}{\prod_{2k\neq b}x_{2k}-x_b} \qquad \,\,\,\,\,\,\,\, \text{$a, b$ are even} \\
\qquad \qquad \qquad \,\,\, 0 \qquad \qquad \qquad \qquad \text{otherwise}.
\end{cases}
\end{align}
**Convention.** For instance, when $a$ is odd, the product $\prod_{2j-1\neq b}(x_{2j-1}-y_a)$ is understood as running through all odd integers from $1$ to $n$, excluding $b$.

The solution to your problem is found by replacing $x_i=i-1$ and $y_j=-j$. Hence, in this case,
\begin{align}
K_{i,j}=\begin{cases}\frac{2(-1)^{a+b}n_1^2}{16^{n_1-1}(2a+2b-3)}
\binom{2n_1+2a-2}{2a-2}\binom{2n_1+2b-2}{2b-2}\binom{2n_1-1}{n_1-a}\binom{2n_1-1}{n_1-b} \qquad i=2a-1,\, j=2b-1 \\
\qquad \frac{(-1)^{a+b}8ab}{16^n(2a+2b-1)}\binom{2n_2+2a}{2a}\binom{2n_2+2b}{2b}\binom{2n_2}{n_2-a}\binom{2n_2}{n_2-b} \qquad \qquad i=2a, \, j=2b \\
\qquad \qquad \qquad \qquad \qquad 0 \qquad \qquad \qquad \qquad \qquad \qquad \text{otherwise}
\end{cases} \end{align}
where we use designating $n_1=\lfloor\frac{n+1}2\rfloor$ and $n_2=\lfloor\frac{n}2\rfloor$.