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 ${-x+j\choose j}{-x-1\choose j}={x+j\choose j}{x-1\choose j}$, we have $P_m(-x)=P_m(x)$. Let $q_j(x)={x\choose 2j}+{-x\choose 2j}.$ Then
\begin{align*}
P_1(x)&=-2q_1(x)+2\\
P_2(x)&=6q_2(x)-6q_1(x)\\
P_3(x)&=24q_3(x)-72q_2(x)+48q_1(x)\\
P_4(x)&=118q_4(x)-704q_3(x)+1522q_2(x)-936q_1(x)\\
P_5(x)&=696q_5(x)-6900q_4(x)+30960q_3(x)-63252q_2(x)+38496q_1(x)\\
P_6(x)&=4824q_6(x)-71640q_5(x)+547572q_4(x)\\
&-2345904q_3(x)+4757916q_2(x)-2892768q_1(x).
\end{align*}
This idea was given by Wilberd van der Kallen.

(3) Let $S_m(x)=xP_m(x)$. Then
$$
S_m(x)=\sum_{i=0}^{m}\sum_{j=0}^{m}{x+j\choose 2j+1}{2j\choose j}{m\choose i}{j\choose i}{i\choose m-j}\frac{3}{(2i-1)(2m-2i-1)}.
$$
Note that ${2j\choose j}{m\choose i}{j\choose i}{i\choose m-j}\frac{1}{(2i-1)(2m-2i-1)}$ is an integer (I can prove this). So the question is equivalent to $$n|S_m(n)$$ for any positive integer $n,m$.

(4) 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 $|n|\le \left\lfloor \frac{m+1}{2} \right\rfloor$.