Example 10 If $0<a<b<c<d<500$, how many ordered quadruples of integers $(a, b, c, d)$ satisfy $a+$ $d=b+c$ and $b c-a d=93$?
From the given conditions, we can set $(a, b, c, d)=(a, a+x, a+y, a+x+y)$, where $x, y$ are integers, and $0<x<y$.
Since $93=b c-a d=(a+x)(a+y)-a(a+x+y)=x y$
and $93=1 \times 93=3 \times 31$, with $x<y$,
we have $(x, y)=(1,93)$ or $(x, y)=(3,31)$.
We will discuss the two cases separately.
(i) When $(x, y)=(1,93)$, $(a, b, c, d)=(a, a+1, a+93, a+94)$, and $a$ can take values $1,2, \cdots, 405$.
(ii) When $(x, y)=(3,31)$, $(a, b, c, d)=(a, a+3, a+31, a+34)$, and $a$ can take values $1,2, \cdots, 465$. There are no duplicates in these two sets of numbers, so there are a total of $405+465=870$ ordered quadruples that satisfy the conditions.