Solution:
Suppose $n = p q r$, where $p < q$ are primes and $r > 1$. Then $p \geq 2$, $q \geq 3$ and $r \geq 2$, not necessarily a prime. Thus we have
$$\begin{aligned} & n-2 \geq n-p = p q r - p \geq 5p > p \\ & n-2 \geq n-q = q(p r - 1) \geq 3q > q \\ & n-2 \geq n-p r = p r(q-1) \geq 2 p r > p r \\ & n-2 \geq n-q r = q r(p-1) \geq q r \end{aligned}$$
Observe that $p, q, p r, q r$ are all distinct. Hence their product divides $(n-2)!$. Thus $n^{2} = p^{2} q^{2} r^{2}$ divides $(n-2)!$ in this case. We conclude that either $n = p q$ where $p, q$ are distinct primes or $n = p^{k}$ for some prime $p$.

Case 1. Suppose $n = p q$ for some primes $p, q$, where $2 < p < q$. Then $p \geq 3$ and $q \geq 5$. In this case
$$\begin{aligned} & n-2 > n-p = p(q-1) \geq 4p \\ & n-2 > n-q = q(p-1) \geq 2q \end{aligned}$$
Thus $p, q, 2p, 2q$ are all distinct numbers in the set $\{1, 2, 3, \ldots, n-2\}$. We see that $n^{2} = p^{2} q^{2}$ divides $(n-2)!$. We conclude that $n = 2q$ for some prime $q \geq 3$. Note that $n-2 = 2q-2 < 2q$ in this case so that $n^{2}$ does not divide $(n-2)!$.

Case 2. Suppose $n = p^{k}$ for some prime $p$. We observe that $p, 2p, 3p, \ldots, (p^{k-1}-1)p$ all lie in the set $\{1, 2, 3, \ldots, n-2\}$. If $p^{k-1}-1 \geq 2k$, then there are at least $2k$ multiples of $p$ in the set $\{1, 2, 3, \ldots, n-2\}$. Hence $n^{2} = p^{2k}$ divides $(n-2)!$. Thus $p^{k-1}-1 < 2k$.

If $k \geq 5$, then $p^{k-1}-1 \geq 2^{k-1}-1 \geq 2k$, which may be proved by an easy induction. Hence $k \leq 4$. If $k=1$, we get $n=p$, a prime. If $k=2$, then $p-1<4$ so that $p=2$ or $3$; we get $n=2^{2}=4$ or $n=3^{2}=9$. For $k=3$, we have $p^{2}-1<6$ giving $p=2$; $n=2^{3}=8$ in this case. Finally, $k=4$ gives $p^{3}-1<8$. Again $p=2$ and $n=2^{4}=16$. However $n^{2}=2^{8}$ divides $14!$ and hence is not a solution.

Thus $n = p, 2p$ for some prime $p$ or $n = 8, 9$. It is easy to verify that these satisfy the conditions of the problem.