There is no solution to this problem. I've a calculus one which makes this problem quite simple. But I don't want to post it. I wish to light this unsolved problem to be solved by non-calculus method.

The shortest solution uses a famous result of George Pólya. If a polynomial takes integer values for all integer arguments, then it is a linear combination with integer coefficients of the polynomials $1,x,\frac {x(x-1)} {2}, \ldots, \frac {x(x-1)\cdots (x-n+1)} {n!}, \ldots$. In our case, $\frac {1} {m} p(x)$ is such, so it can be written as $\alpha_k + \alpha_{k-1}x + \alpha_{k-2} \frac {x(x-1)} {2} + \cdots + \alpha_{0} \frac {x(x-1)\cdots (x-k+1)} {k!}$. Equalling the dominant coefficients we get $\frac {1} {m} a_0 = \alpha_0 \frac {1} {k!}$, so $m\alpha_0 = k!a_0$, hence $m \mid k!a_0$. Conversely, the polynomial $p(x) = a_0x(x-1)\cdots (x-k+1)$ is a good example, since $k! \mid x(x-1)\cdots (x-k+1)$ for all integers $x$, therefore $m \mid k!a_0 \mid p(x)$.