$6 \cdot 66$ Find all ordered pairs of positive integers $(m, n)$ such that $\frac{n^{3}+1}{m n-1}$ is an integer.
(36th International Mathematical Olympiad, 1994)
[Solution] Since $\left(m n-1, m^{3}\right)=1$, then $m n-1 \mid n^{3}+1$ is equivalent to $m n-1 \mid m^{3}\left(n^{3}+1\right)$. Clearly,
$$
m^{3}\left(n^{3}+1\right)=\left(m^{3} n^{3}-1\right)+\left(m^{3}+1\right),
$$

Since $m^{3} n^{3}-1=(m n-1)\left(m^{2} n^{2}+m n+1\right)$,
thus $m n-1 \mid m^{3} n^{3}-1$,
therefore, $m n-1 \mid m^{3}+1$.
From this, we know that $m n-1$ divides $n^{3}+1$ is equivalent to $m n-1$ divides $m^{3}+1$.
If $m=n$, then $\frac{n^{3}+1}{m n-1}=\frac{n^{3}+1}{n^{2}-1}=\frac{n^{2}-n+1}{n-1}=n+\frac{1}{n-1}$,
by the problem's condition, $\frac{1}{n-1}$ must be an integer. Thus, $n-1$ can be $\pm 1$, and since $n$ is a positive integer, then $n=2$.
Therefore, $(m, n)=(2,2)$ is a solution.
If $m \neq n$, without loss of generality, assume $m>n$.
If $n=1$, then $\frac{n^{3}+1}{m-1}=\frac{2}{m-1}$,
thus, by $\frac{2}{m-1}$ being an integer, we get $m=2$, or $m=3$.
Thus, we have the solutions $(m, n)=(2,1),(3,1)$.
By symmetry, we also have $(m, n)=(1,2),(1,3)$.
If $n \geqslant 2$, then $n^{3}+1 \equiv 1 \quad(\bmod n)$, and $m n-1 \equiv-1 \quad(\bmod n)$.
Therefore, there exists some positive integer $k$, such that $\frac{n^{3}+1}{m n-1}=k n-1$.
Since $m>n$, then $k n-1=\frac{n^{3}+1}{m n-1}<\frac{n^{3}+1}{n^{2}-1}=n+\frac{1}{n-1}$,
so $(k-1) n<1+\frac{1}{n-1}$.

Since $n \geqslant 2$, then $\frac{1}{n-1} \leqslant 1$, thus we can only have $k=1$.
$$
\frac{n^{3}+1}{m n-1}=k n-1=n-1.
$$

That is, $n^{3}+1=(m n-1)(n-1)=m n^{2}-n-m n+1$,
simplifying to $\quad n^{2}=m n-1-m \quad m=\frac{n^{2}+1}{n-1}=n+1+\frac{2}{n-1}$.
By $\frac{2}{n-1}$ being an integer and $n$ being a positive integer, we get $n=2$ or $n=3$. Correspondingly, $m=5$.
Thus, we have the solutions $(m, n)=(5,2),(5,3)$,
and by symmetry, $(m, n)=(2,5),(3,5)$.
In summary, there are nine solutions $(m, n)=(2,2),(2,1),(3,1),(5,2),(5,3),(1,2)$, $(1,3),(2,5),(3,5)$.