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.

I list the polynomials when $m=1,2,3,4,5$ \begin{align*} P_1(x)&=-2(x^2-1),\\ P_2(x)&=\frac{x^2(x^2-1)}{2},\\ P_3(x)&=\frac{x^2(x^2-1)(x^2-4)}{3\cdot 5},\\ P_4(x)&=\frac{x^2(x^2-1)(x^2-4)(59x^2-419)}{2^5\cdot 3^2 \cdot 5 \cdot 7},\\ P_5(x)&=\frac{x^2(x^2-1)(x^2-4)(x^2-9)(29x^2-239)}{2^4\cdot 3^3\cdot 5^2\cdot 7}. \end{align*} Some remarks:

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

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

(3) 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}-8{x\choose 2}-{x\choose 1},\\ 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*}