Suppose that there exist only finitely many such $n$. Hence, there exists a positive integer constant $t$, such that for all $n \geq t+1$, $f(n)$ is divisible by some $f(i)$ for $i \in \{2, 3,\ldots, n-1\}$. Let $n \equiv 1 \pmod {f(2)f(3)\ldots f(t)}$; this means that $f(n) \equiv f(1) \pmod {f(2)f(3)\ldots f(t)}$. If there exists a positive integer $k \in [2;t]$ such that $f(k) \mid f(n)$, this yields that $f(k) \mid f(1)$, which is impossible for size reasons since the polynomial has non-negative integer coefficients. Hence, $f(k) \nmid f(n)$ for all positive integers $k \in [2; t]$. Due to our assumption, there exists a positive integer $t<a_1<n$, such that $f(a_1) \mid f(n)$. Applying this again, there exists a positive integer $a_2<a_1$, such that $f(a_2) \mid f(a_1)$, which means that $a_2>t$, otherwise there is a $k \in [2; t]$ such that $f(k) \mid f(a_1) \mid f(n)$. Continuing in the same manner, we obtain a strictly decreasing sequence $\{a_n\}$ of positive integers bounded in $[t+1; n-1]$, which is a contradiction, so we are done.