Find the inverse of a matrix that is very similar to the Hilbert matrix
As observed in comments, the problem is equivalent to finding the inverse of the matrices $$H_\lambda:=\Big[{1\over i+j+\lambda}\Big]_{1\le i\le m\atop 1\le j\le m},$$ for $\lambda=-1/2$ and $\lambda=-3/2$ and order $m=\lceil n/2\rceil$, respectively $m=\lfloor n/2\rfloor$. These are also particular cases of Cauchy matrices, whose inverses admit the explicit Schetcher's formula quoted in the linked question (see the wiki article on Cauchy matrices and the various answers in the linked question). As a matter of fact, in this particular case the inversion formula takes quite a simpler form, and it turns out that the entries of $H_\lambda^{-1}$ are products of linear factors $\lambda+k$, with $2\le k\le 2m$ and multiplicity not larger than $2$. Precisely
$$H^{-1}_\lambda={1\over(m-1)!^2}\bigg[(-1)^{i+j}{m-1\choose i-1}{m-1\choose j-1}(\lambda+i+1)^{\overline m}(\lambda+j+1)^{\overline m}(\lambda+i+j)^{-1}\bigg]_{1\le i\le m\atop 1\le j\le m}$$ where $x^{\overline m}:=x(x+1)\dots(x+m-1)$ denotes a rising factorial.
We 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 $n_2=\lfloor\frac{n}2\rfloor$.
These matrices have been considered here.