Notice that, if $p$ is a prime and $g$ is a polynomial with integer coefficients such that $g(n) \equiv 0(\bmod p)$ for some $n$, then $g(n+m p)$ is divisible by $p$ as well for any integer multiple $m p$ of $p$. Therefore, it suffices to find the smallest possible degree of a polynomial $f$ for which $f(0), f(1), f(2), \ldots, f(16)$ are divisible by 17 and by 5. There is a polynomial of degree 17 with integer coefficients having $f(0)=f(1)=\cdots=f(16)=0$, namely $f(x)=(x)(x-1)(x-2) \cdots(x-16)$. Thus the minimal degree is no larger than 17. Now, let $f$ be such a polynomial and consider $f$ modulo 17. The polynomial has 17 roots, so it must be at least degree 17 when taken modulo 17. Thus $f$ has degree at least 17 as well.