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 $S_m(x)$ is a polynomial with $\mathbb Z$-coefficients provided ${2j\choose j}{m\choose i}{j\choose i}{i\choose m-j}\frac{1}{(2i-1)(2m-2i-1)}$ 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 $|n|\le \left\lfloor \frac{m+1}{2} \right\rfloor$.