Example 8 Find all positive integers $a, b, c$ such that $1<a<b<c$, and $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.
(33rd IMO Problem)
Let $a=1+t_{1}, b=1+t_{2}, c=1+t_{3}$, then $1 \leqslant t_{1}<t_{2}<t_{3}$. Also,
$$
\begin{aligned}
2 & \leqslant \frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)-1}{t_{1} t_{2} t_{3}} \\
& <\frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)}{t_{1} t_{2} t_{3}} \\
& =\left(1+\frac{1}{t_{1}}\right)\left(1+\frac{1}{t_{2}}\right)\left(1+\frac{1}{t_{3}}\right) \\
& \leqslant\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=4,
\end{aligned}
$$

Therefore, $\frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)-1}{t_{1} t_{2} t_{3}}$ is either 2 or 3.
(1) If $\frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)-1}{t_{1} t_{2} t_{3}}=2$, then $\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}+\frac{1}{t_{1} t_{2}}+\frac{1}{t_{2} t_{3}}+\frac{1}{t_{3} t_{1}}=1$.
Clearly, $t_{1} \neq 1$.

If $t_{1} \geqslant 3$, then $t_{2} \geqslant 4, t_{3} \geqslant 5$, at this time
the left side of (1) $\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12}+\frac{1}{20}+\frac{1}{15}=\frac{59}{60}<1$.
Thus, $t_{1}$ can only be 2, at this time (1) becomes
$$
\frac{1}{t_{2}}+\frac{1}{t_{3}}+\frac{1}{2 t_{2}}+\frac{1}{t_{2} t_{3}}+\frac{1}{2 t_{3}}=\frac{1}{2}.
$$

It is easy to see that $3 \leqslant t_{2} \leqslant 5$. Only when $t_{2}=4$, $t_{3}=14$ is an integer.
Therefore, (1) has only one solution $(2,4,14)$.
(2) If $\frac{\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right)-1}{t_{1} t_{2} t_{3}}=3$,
then $\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}+\frac{1}{t_{1} t_{2}}+\frac{1}{t_{2} t_{3}}+\frac{1}{t_{3} t_{1}}=2$.
Thus, $t_{1}=1$, and (2) becomes $\frac{2}{t_{2}}+\frac{2}{t_{3}}+\frac{1}{t_{2} t_{3}}=1$.
It is easy to see that $2 \leqslant t_{2} \leqslant 3$. Only when $t_{2}=3$, $t_{3}=7$ is an integer.
Therefore, (2) has only the positive integer solution $(1,3,7)$.
In summary, the positive integers $(a, b, c)$ that satisfy the problem are $(3,5,15)$ and $(2,4,8)$.