Entries of the inverse of $\left[\frac{1}{x+i+j-1}\right]_{i,j\in\{1,2,\ldots,n\}}$ are polynomials in $x$. Let $n$ be a positive integer.  Define $$\textbf{A}_n(x):= \left[\frac{1}{x+i+j-1}\right]_{i,j\in\{1,2,\ldots,n\}}$$ as a matrix over the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in variable $x$.  
(a) Prove that the Hilbert matrix $\textbf{A}_n(0)$ is an invertible matrix over $\mathbb{Q}$ and all entries of the inverse of $\textbf{A}_n(0)$ are integers.
(b) Determine the greatest common divisor (over $\mathbb{Z}$) of all the entries of $\big(\textbf{A}_n(0)\big)^{-1}$.
(c) Show that $\textbf{A}_n(x)$ is an invertible matrix over $\mathbb{Q}(x)$ and every entry of the inverse of $\textbf{A}_n(x)$ is a polynomial in $x$.
(d) Prove that $x+n$ is the greatest common divisor (over $\mathbb{Q}[x]$) of all the entries of $\big(\textbf{A}_n(x)\big)^{-1}$.
Parts (a) and (c) are known. Parts (b) and (d) are open. Now, Part (d) is known (see i707107's solution below), but Part (b) remains open, although it seems like the answer is $n$.

Recall that 
$$\binom{t}{r}=\frac{t(t-1)(t-2)\cdots(t-r+1)}{r!}$$
for all $t\in\mathbb{Q}(x)$ and $r=0,1,2,\ldots$.  According to i707107, the $(i,j)$-entry of $\big(\textbf{A}_n(x)\big)^{-1}$ is given by
$$\alpha_{i,j}(x)=(-1)^{i+j}\,(x+n)\,\binom{x+n+i-1}{i-1}\,\binom{x+n-1}{n-j}\,\binom{x+n+j-1}{n-i}\,\binom{x+i+j-2}{j-1}\,.\tag{*}$$
This means that, for all integers $k$ such that $k\notin\{-1,-2,\ldots,-2n+1\}$, the entries of $\big(\textbf{A}_n(k)\big)^{-1}$ are integers.  I now have a new conjecture, which is the primary target for the bounty award.

Conjecture:  The greatest common divisor $\gamma_n(k)$ over $\mathbb{Z}$ of the entries of $\big(\textbf{A}_n(k)\big)^{-1}$, where $k$ is an integer not belonging in the set $\{-1,-2,\ldots,-2n+1\}$, is given by $$\gamma_n(k)=\mathrm{lcm}(n,n+k)\,.$$

It is clear from (*) that $n+k$ must divide $\gamma_n(k)$.  However, it is not yet clear to me why $n$ should divide $\gamma_n(k)$.  I would like to have a proof of this conjecture, or at least a proof that $n \mid \gamma_n(k)$.

Let $M_n$ denote the (unitary) cyclic $\mathbb{Z}[x]$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)$.  Then, the (unitary)  $\mathbb{Z}[x]$-module $N_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}[x]$-submodule of $M_n$.  
We also denote by $\tilde{M}_n$ for the (unitary)  $\mathbb{Z}$-module generated by $\dfrac{1}{\big((n-1)!\big)^2}\,(x+n)\,x^l$ for $l=0,1,2,\ldots,2n-2$.  Then, the (unitary) $\mathbb{Z}$-module $\tilde{N}_n$ generated by the entries of $\big(\textbf{A}_n(x)\big)^{-1}$ is a $\mathbb{Z}$-submodule of $\tilde{M}_n$.
For example, $M_2/N_2$ is isomorphic to the (unitary) $\mathbb{Z}[x]$-module $\mathbb{Z}/2\mathbb{Z}$ (in which $x$ acts trivially), and $\tilde{M}_2/\tilde{N}_2$ is isomorphic to the (unitary) $\mathbb{Z}$-module $\mathbb{Z}/2\mathbb{Z}$.  Hence, $\left|M_2/N_2\right|=2=\left|\tilde{M}_2/\tilde{N}_2\right|$.  For $n=3$, Mathematica yields
$$\tilde{M}_3/\tilde{N}_3\cong (\mathbb{Z}/2\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/4\mathbb{Z})^{\oplus 3}\,,$$
as abelian groups.  That is, $\left|\tilde{M}_3/\tilde{N}_3\right|=1152$.  On the other hand, 
$$M_3/N_3\cong \mathbb{Z}[x] \big/\left(12,2x^2+6x+4,x^4-x^2\right)$$
as $\mathbb{Z}[x]$-modules, which gives $\left|M_3/N_3\right|=576$.    

Question:  Describe the factor $\mathbb{Z}[x]$-module $M_n/N_n$ and the factor $\mathbb{Z}$-module $\tilde{M}_n/\tilde{N}_n$.  It is easily seen that $\left|M_n/N_n\right|\leq\left|\tilde{M}_n/\tilde{N}_n\right|$.  What are $\left|M_n/N_n\right|$ and $\left|\tilde{M}_n/\tilde{N}_n\right|$?  It can be shown also that the ratio $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ is an integer, provided that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite.  Compute $\dfrac{\left|\tilde{M}_n/\tilde{N}_n\right|}{\left|M_n/N_n\right|}$ for all integers $n>0$ such that $\left|\tilde{M}_n/\tilde{N}_n\right|<\infty$.  Is it always the case that $\left|\tilde{M}_n/\tilde{N}_n\right|$ is finite?

Apart from the conjecture above, this question is also eligible for the bounty award.  I have not yet fully tried to deal with any case involving $n>3$.  However, for $n=4$, the module $\tilde{M}_4/\tilde{N}_4$ is huge:
$$  \tilde{M}_4/\tilde{N}_4\cong (\mathbb{Z}/2\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/3\mathbb{Z})^{\oplus 3}\oplus(\mathbb{Z}/8\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/9\mathbb{Z})^{\oplus 2}\oplus(\mathbb{Z}/16\mathbb{Z})\oplus(\mathbb{Z}/27\mathbb{Z})$$
as abelian groups.
 A: This is a solution to (d), and partially for (b). 
We use Cauchy Matrix, and its evaluation of inverse given by Schechter: 
If $T$ is a $n\times n$ Cauchy matrix on the sequences $\{x_i\}$, $\{y_j\}$, then
$S=T^{-1}=[s_{ij}]$ is given by:
$$s_{ij} = (x_j - y_i) A_j(y_i) B_i(x_j) $$
where
$$A_i(t) = \frac{A(t)}{A^\prime(x_i)(t-x_i)} \quad\text{and}\quad B_i(t) = \frac{B(t)}{B^\prime(y_i)(t-y_i)}$$
with
$$A(t) = \prod_{i=1}^n (t-x_i) \quad\text{and}\quad B(t) = \prod_{i=1}^n (t-y_i). $$
The matrix $\mathbf{A}_n(x)$ is considered as a Cauchy matrix on the sequences 
$$x_i = x+i  \quad\text{and} \quad y_j = -j+1.$$
Using the above inverse formula, the $ij$ entry of the inverse up to sign is 
$$
\frac{ (x+n)\cdot (x+i) \cdots (x+n-1)}{(n-i)!} \frac{ (x+n+1)\cdots (x+n+i-1)}{(i-1)!} \frac{ (x+j) \cdots (x+j+n-1)}{(x+i+j-1)(n-j)!(j-1)!}
$$
We see that every entry is divisible by $x+n$. 
To prove that the GCD is $x+n$, we need to prove that any possible factor other than $x+n$ which is present in one entry is not present in some another entry. This is easy to see.
Moreover, the factor $(x+n)^2$ is present for most of times, but it is possible to have only one $x+n$ in the factorization (in case $x+i+j-1= x+n$).
For (b), we use the well known formula for inverse of Hilbert matrix:
$$
(H^{-1})_{ij}=(-1)^{i+j}(i+j-1){n+i-1 \choose n-j}{n+j-1 \choose n-i}{i+j-2 \choose i-1}^2
$$
Upon rearranging, we obtain that the $ij$ entry up to sign is
$$
n\frac{(n+i-1)\cdots (n+1)}{(i-1)!} \frac{(n-1)\cdots (n-j+1)}{(j-1)!} \binom{n+j-1}{i+j-1} \binom{i+j-2}{i-1}
$$
which is simplified to
$$
n \binom{n+i-1}{i-1} \binom{n-1}{j-1} \binom{n+j-1}{i+j-1} \binom{i+j-2}{i-1}.
$$
This is clearly divisible by $n$, but I was not able to show that $n$ is the GCD which I conjecture to be. 
A: For Part (b), according to i707107's answer, the $(i,j)$-entry of $\textbf{H}_n:=\big(\textbf{A}_n(0)\big)^{-1}$ is given by $$h_{i,j}:=(-1)^{i+j}\,n\,\binom{n+i-1}{i-1}\,\binom{n-1}{j-1}\,\binom{n+j-1}{i+j-1}\,\binom{i+j-2}{i-1}\,.$$
Hence, $n$ is a divisor of the greatest common divisor $g_n$ over $\mathbb{Z}$ of the entries of $\textbf{H}_n$.
Note that
$$\left|h_{1,j}\right|=n\,\binom{n-1}{j-1}\,\binom{n+j-1}{j}=n\,\binom{n+j-1}{n-1}\,\binom{n-1}{j-1}=n\,\binom{n+j-1}{j-1}\,\binom{n}{j}\,;$$
in particular,
$$h_{1,1}=n^2\,.$$
Ergo, $$n\mid g_n\mid n^2\,.$$
If $p$ is a prime divisor of $n$ such that $p^k$ is the largest power of $p$ that divides $n$, then using Lucas's Theorem, we know that 
$$\binom{n+p^k-1}{p^k-1}\equiv 1\pmod{p}$$
and
$$\binom{n}{p^k}\equiv \frac{n}{p^k}\pmod{p}\,.$$
Therefore, $p$ does not divide $\dfrac{h_{1,p^k}}{n}$, whence $p\nmid \dfrac{g_n}{n}$.  Hence, the greatest common divisor of the entries of $\textbf{H}_n=\big(\textbf{A}_n(0)\big)^{-1}$ must be
$$g_n=n\,.$$
