Example 18 (Problem 1494 from "Mathematics Bulletin") Try to find all positive integers $a, b, c, d$, where $1<a<b<c<d$, such that $a b c d-1$ is an integer multiple of $(a-1)(b-1)(c-1)(d-1)$.
Let $k=\frac{a b c d-1}{(a-1)(b-1)(c-1)(d-1)} \in \mathbf{Z}^{+}$. It is easy to see that $a, b, c, d$ are all odd or all even, and $11)$.
If $a \geqslant 5$, then $1<k<\frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} \cdot \frac{8}{7}=2$, $k$ does not exist;
If $a=4$, then $b, c, d$ are all even, $b \geqslant 6, c \geqslant 8, d \geqslant 10$, thus $k$ is odd, and
$3 \leqslant k<\frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdot \frac{10}{9}=\frac{128}{63}<3$, $k$ does not exist;
If $a=3$, then $b, c, d$ are all odd, $b \geqslant 5, c \geqslant 7, d \geqslant 9$, thus $1<k<\frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdot \frac{9}{8}=\frac{315}{128}<3$,
so $k=2$.
If $b=7$, because $3 \times 3 b c d-1$,
so $3 \times(a-1)(b-1)(c-1)(d-1)=2 \cdot b \cdot(c-1)(d-1)$, a contradiction.
Therefore, $b \neq 7$. Consequently, $b, c, d \neq 1(\bmod 3)$.
If $b \geqslant 9$, then $2=k<\frac{3}{2} \cdot \frac{9}{8} \cdot \frac{11}{10} \cdot \frac{15}{14}=\frac{891}{448}<2$, a contradiction.
If $b=5$, then $2 \cdot 2 \cdot 4(c-1)(d-1)=15 c d-1$, i.e., $(c-16)(d-16)=239$ (prime), so $a=3, b=5$, $c=17, d=255$.
If $a=2$, similarly, we get $k=3$, then $3 \mid a b c d-1$, thus $3 \mid a b c d$.
If $b \neq 4$, then $b \geqslant 8, c \geqslant 10, d \geqslant 14$, thus $1<k<2 \cdot \frac{8}{7} \cdot \frac{10}{9} \cdot \frac{14}{13}=\frac{2240}{819}<3$, a contradiction, so $b=4$, from $3 \cdot 1 \cdot$ $3 \cdot(c-1)(d-1)=2 \cdot 4 c d-1$, we get $(c-9)(d-9)=71$ (prime).
So $a=2, b=4, c=10, d=80$.
In summary, the solutions are $(3,5,17,255),(2,4,10,80)$.
Note: This problem is a generalization of Problem 1 from the 33rd IMO.
In solving the problem, making assumptions about multiple scenarios for a certain object is a method of specific conjecture in the problem—classification and discussion.