Let $m$ be any positive integer.
$$
P_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}{x+j\choose j}{x-1\choose j}{j\choose i}{m\choose i}{i\choose m-j}\frac{3}{(2i-1)(2j+1)(2m-2i-1)}.
$$
Question: $P_m(x)$ always has integer values at all integers.
 
Some remarks:

(1)  A polynomial is $\mathbb Z$-valued iff its unique expansion in basis $\left\{{x\choose k}\right\}$ has $\mathbb Z$-coefficients. This idea was Allen Knutson's.
For $m=1,2,3,4$, I list the polynomials in this basis, which is more clear that $P_m(x)$ is $\mathbb Z$-valued.
\begin{align*}
P_1(x)&=-4{x\choose 2}-2{x\choose 1}+2,\\
P_2(x)&=12{x\choose 4}+18{x\choose 3}+6{x\choose 2},\\
P_3(x)&=48{x\choose 6}+120{x\choose 5}+96{x\choose 4}+24{x\choose 3},\\
P_4(x)&=236{x\choose 8}+826{x\choose 7}+1070{x\choose 6}+610{x\choose
5}+134{x\choose 4}+4{x\choose 3}
\end{align*}

(2) Since ${-x+j\choose j}{-x-1\choose j}={x+j\choose j}{x-1\choose j}$. Then $P_m(-x)=P_m(x)$.

(3) It is easy to see that letting $|x|\le \left\lfloor \frac{m+1}{2} \right\rfloor$ be a integer, we have
$$
{x+j\choose j}{x-1\choose j}{j\choose i}{m\choose i}{i\choose m-j}=0.
$$
Then $P_m(n)=0$ for any integer $|n|\le \left\lfloor \frac{m+1}{2} \right\rfloor$.