Main idea:find primes $p,q$ and $n$ such that $pq|n^2+n+a$ and $n<pq$.

We will construct an infinity of numbers in which the requirement of the problem is verified. Consider $n+1-a=p^2$.So $n^2+n+a=(n+1)^2-n-1+a=(p^2+p+a)\cdot(p^2-p+a)$. It is clear that for sufficiently large p we have $a<\frac{p^2+p+a}{a}<\frac{p^2+p+a}{2}<p^2-p+a$ . If $a>1$ we take $p\equiv-1 \pmod a$ sufficiently large so $p^2+p+a\equiv 0 \pmod a$ and $1<a<\frac{p^2+p+a}{a}<p^2-p+a<p^2+a-1=n$ so $(p^2+a-1)!$ is divisible by $ (p^2+p+a)\cdot(p^2-p+a)$ for infinitely many p.So $n!$ is divisible by $n^2 + n + a$ for infinitely many positive integers $n.$. If $a=1$ we have that $n=p^2$.We take $ p \equiv 1 \pmod 3 $ sufficiently large so $1<3<\frac{p^2+p+1}{3}<p^2-p+1<p^2=n$ so $(p^2)!$ is divisible by $ (p^2+p+1)\cdot(p^2-p+1)$ for infinitely many p.So $n!$ is divisible by $n^2 + n + 1$ for infinitely many positive integers $n.$