We shall find all natural numbers $n$ s.t. $(1) \;\; (2n)! \equiv 0 \pmod{n! + 2}$.. By inspection we find that (1) for $n \leq 3$ is satisfies only if $n \in \{2,3\}$. Assume $n \geq 4$ satisfies (1). Then ${\textstyle GCD(n!, \frac{n!}{2} + 1) = 1}$, which combined with (1) and the fact that $(n!)^2 | (2n)!$ implies $\frac{(2n)!}{(n!)^2(\frac{n!}{2} + 1)} = \frac{2(2n)!}{(n!)^2(n! + 2)} = k \in \mathbb{N}$, i.e. $(2)\;\; 2{2n \choose n} = k(n! + 2)$. The fact that $n \geq 4$ and ${2n \choose n}$ is even combined with (2) give us $2|k$. Therefore $k \geq 2$, which according to (2) means ${2n \choose n} \geq n! + 2 > n!$, yielding $(3) \;\; \frac{(2n)!}{(n!)^3} = f(n) > 1$. Now $f(n+1) = \frac{(2n+2)!}{[(n+1)!]^3} = \frac{(2n+2)(2n+1)(2n)!}{(n+1)^3 \cdot (n!)^3} = \frac{2(2n+1)}{(n+1)^2} f(n)$, which (since $n \geq 4$) give us $f(n+1) < f(n)$. Seeing that $f(7)<1$, we obtain (according to (3)) $n \leq 6$. Checking (1) for $n=4,5,6$, we find that none of these three values of $n$ satisfies (1). Conclusion: The only natural numbers $n$ satisfying (1) are $n=2$ and $n=3$.