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.