Problem 1561

Without loss of generality one may assume that $f(1) \in \mathbb{Z}$. Since for all positive $k$, we have $\frac{f(k+1)}{f(1)}$ is integer, then we conclude that on all positive integer points our polynomial gets integer values. Assume that $\operatorname{deg}(f)=d$ then, according to Lagrange interpolation formula we get
$$f(x)=\sum_{i=1}^{d+1} f(i) \frac{(x-1) \ldots(x-(i-1))(x-(i+1)) \ldots(x-(d+1))}{(i-1) \ldots 1(-1) \ldots(i-(d+1))},$$
and all numbers are rational, so $p(x)$ is a polynomial with rational coefficients.
By multiplying by a constant we can get $f(x)$ with integer coefficients. If $f(0)=0$ then we are done. Assume that $f(0) \neq 0$. Let's fix positive integer $k$ and denote $f(n)=f(n+1) f(n+2) \ldots f(n+k)$ and $M=f(0)$.
According to the problem condition, we have $\frac{f(n)}{M} \in \mathbb{Z}$ for all positive integer $n$. Since $f(x)$ is polynomial with integer coefficients, then
$$f(-1) \equiv f(2|M|-1) \equiv 0 \quad(\bmod M),$$
which means
$$\frac{f(-1)}{M}=\frac{f(0)}{f(k)} \in \mathbb{Z},$$
for all positive integers $k$. It means $f(0)=0$ or $f(x)$ is divisible by $x$. $\square$