Example 6 How many ordered quadruples of integers $(a$, $b, c, d)$ satisfy $0<a<b<c<d<500, a+d=b$ $+c$ and $bc-ad=93$?
(11th American Invitational Mathematics Examination)
Solution: Let $b=a+t, t \in N$, then $c=d-t$. Substituting into $b c-a d=93$, we get $(d-a-t) t=3 \times 31$.

When $t=1$, we have $b=a+1, c=a+93, d=$ $a+94$. Since $a+94<500$, we have $0<a<406$. Thus, there are 405 sets of $(a, b, c, d)$.

When $t=3$, we have $b=a+3, c=a+31, d=$ $a+34$. In this case, $0<a<466$, so there are 465 sets of $(a, b, c, d)$.

When $t=31,93$, it is easy to see that there are no sets of $(a, b, c, d)$ that satisfy the conditions.

Therefore, there are a total of $405+465=870$ sets of integers $(a, b, c, d)$ that satisfy the conditions.