Let's generalize, and also give explicit solutions.

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 $K:=M^{-1}$ has entries given by
\begin{align}
K_{a,b}=\begin{cases}
\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} \\
\,\,\,\,\,\,\,\, \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}
The solution to your problem is found by replacing
$$x_i=i+1 \qquad \text{and} \qquad y_j=j.$$
**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 $a$.