7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$
7.870.

Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$

where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$

Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$

where $a=1,2, \cdots, 405$;
Second case
$$
(a, b, c, d)=(a, a+3, a+31, a+34) \text {, }
$$

where $a=1,2, \cdots, 465$.
Thus, the total number of integer quadruples that satisfy the condition is
$$
405+465=870 \text { (sets). }
$$