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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 $H_\lambda$ entries 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 $\overline m$ denote a rising factorial.

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.

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 $H_\lambda$ entries are products of linear factors $\lambda+k$, with $2\le k\le 2m$ and multiplicity not larger than $2$. Precisely

$$H^{-1}_\lambda:={1\over(n-1)!^2}\bigg[(-1)^{i+j}{m-1\choose i-1}{m-1\choose j-1}(\lambda+i+1)^{\overline m}(\lambda+j+1)^{\overline m}(\lambda+i+j)^{-1}\bigg]_{1\le i\le m\atop 1\le j\le m}$$ where $\overline m$ denote a rising factorial.

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 $H_\lambda$ entries 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 $\overline m$ denote a rising factorial.

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 $H_\lambda$ entries are products of linear factors $\lambda+k$, with $2\le k\le 2m$ and multiplicity not larger than $2$. Precisely

$$H^{-1}_\lambda:={1\over(n-1)!^2}\bigg[(-1)^{i+j}{m-1\choose i-1}{m-1\choose j-1}(\lambda+i+1)^{\overline m}(\lambda+j+1)^{\overline m}(\lambda+i+j)^{-1}\bigg]_{1\le i\le m\atop 1\le j\le m},$$ where $\overline m$ denote a rising factorial.

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 $H_\lambda$ entries are products of linear factors $\lambda+k$, with $2\le k\le 2m$ and multiplicity not larger than $2$. Precisely

$$H^{-1}_\lambda:={1\over(n-1)!^2}\bigg[(-1)^{i+j}{m-1\choose i-1}{m-1\choose j-1}(\lambda+i+1)^{\overline m}(\lambda+j+1)^{\overline m}(\lambda+i+j)^{-1}\bigg]_{1\le i\le m\atop 1\le j\le m}$$ where $\overline m$ denote a rising factorial.