Example 4 (2004 Slovenia Mathematical Olympiad Selection Test) Find all positive integer solutions $a, b, c$ for which the expression $\left(b-\frac{1}{a}\right)\left(c-\frac{1}{b}\right)\left(a-\frac{1}{c}\right)$ is an integer.
Due to the symmetry of the problem, we consider the case $a \leqslant b \leqslant c$. Noting that the value of the expression is an integer, we have $a b c \mid[(a b-1)(b c-1)(c a-1)]$,
which means $a b c \mid\left(a^{2} b^{2} c^{2}-a^{2} b c-a b^{2} c-a b c^{2}+a b+b c+a c-1\right)$,
thus $a b c \mid(a b+b c+c a-1)$,
so $a b c \leqslant a b+b c+c a-1<a b+b c+c a \leqslant c(a+a+b)$,
hence $a b<2 a+b \leqslant 3 b$. Therefore, $a \leqslant 3$.
If $a=1$, then $b c \mid(b+c-1)$, and $b c-b-c+1=(b-1)(c-1) \leqslant 0$,
which necessarily leads to $b=1$ (or $c=1$, but $b \leqslant c$).
If $a=2$, then $2 b c \mid(b c+2 b+2 c-1)$.
Thus $b c \leqslant 2 b+2 c-1<2 b+2 c \leqslant 4 c$, hence $b<4$.
If we take $b=2$, then $4 c \mid 4 c+3$.
Thus $4 c \mid 3$, which is impossible.
If we take $b=3$, then $6 c \mid 5 c+5$, so $6 c \leqslant 5 c+5 \Rightarrow c \leqslant 5$.
By checking all cases, we find that only $c=5$ is possible.
Therefore, all solutions (with $a \leqslant b \leqslant c$) are $(1,1, c)$ and $(2,3,5)$, where $c$ is arbitrary. Due to the symmetry of the problem, the order of $a, b, c$ is arbitrary, so the solutions are
$$
(1,1, c),(1, c, 1),(c, 1,1),(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2) \text {. }
$$