$3 \cdot 10$ Find all odd natural numbers $n$ such that $(n-1)!$ is not divisible by $n^2$.
(4th All-Russian Mathematical Olympiad, 1964)
[Solution] First, we can prove that when $n$ is an odd prime and $n=9$, $(n-1)!$ cannot be divisible by $n^{2}$.

When $n$ is an odd prime, then $(n-1)!$ does not contain the factor $n$, so $(n-1)!$ cannot be divisible by $n^{2}$.
When $n=9$, it is clear that 8! is divisible by 9, but not by $9^{2}=81$.
Next, we prove that when $n$ is an odd composite number and $n \neq 9$, $(n-1)!$ must be divisible by $n^{2}$.
If $n$ is an odd composite number and not the square of an odd prime.
Let $n=a b, \quad a \geqslant 3, \quad b \geqslant 3, \quad a \neq b$.
In this case, $a, 2 a, b, 2 b$ must be factors of the product $1 \cdot 2 \cdot 3 \cdots(n-1)$, so $(n-1)!$ can be divisible by $a^{2} b^{2}=n^{2}$.
If $n$ is the square of an odd prime that is at least 5.
Let $n=p^{2}, \quad p$ is a prime, $p \geqslant 5$.
In this case, we must have $\quad p^{2}-1 \geqslant 4 p$.
Therefore, in $(n-1)!=\left(p^{2}-1\right)!$ there must be factors $p, 2 p, 3 p, 4 p$. Thus, $(n-1)!$ can be divisible by $\left(p^{2}\right)^{2}=p^{4}$.
In summary, the odd natural numbers that meet the requirements of the problem are $n$ being an odd prime and $n=9$.