7. Let $a, b$ be positive integers, the quotient when $a^{2}+b^{2}$ is divided by $a+b$ is $q$, and the remainder is $r$, and $q^{2}+r=2010$. Find the value of $a b$.


7. Let $a \leqslant b$.

From $a^{2}+b^{2}>b^{2}-a^{2}=(b-a)(b+a)$, we get $q \geqslant b-a$.
Also, from $a^{2}+b^{2} \geqslant a^{2}+a b=a(a+b)$, we get $q \geqslant a$.
By the properties of division with remainder, we have
$0 \leqslant r<a+b$.
Thus, $0 \leqslant r<a+b=a+a+(b-a) \leqslant 3 q$.
From $q^{2}+r=2010$, we get
$q^{2} \leqslant 2010<q^{2}+3 q$.
The positive integer solution to this inequality is only $q=44$, at which time, $r=74$.
Then, $a^{2}+b^{2}=44(a+b)+74$,
$(a-22)^{2}+(b-22)^{2}=74+2 \times 22^{2}=1042$.
Let $x=\min \{|a-22|,|b-22|\}$,
$y=\max \{|a-22|,|b-22|\}$.
Then $x^{2}+y^{2}=1042 \equiv 2(\bmod 8)$.
Thus, $x$ and $y$ are both positive odd numbers.
From $x^{2} \leqslant y^{2}$, we get
$y^{2} \leqslant x^{2}+y^{2}=1042 \leqslant 2 y^{2}$,

i.e., $23 \leqslant y \leqslant 32$.
Therefore, the possible values of $y$ are $23, 25, 27, 29, 31$, a total of five.

When $y=23,25,27,29$, $x^{2}=513,417,313,201$, none of which have positive integer solutions.
When $y=31$, $x^{2}=81 \Rightarrow x=9$.
Thus, $\left\{\begin{array}{l}|a-22|=31, \\ |b-22|=9\end{array}\right.$ has no solution;
$\left\{\begin{array}{l}|a-22|=9, \\ |b-22|=31\end{array}\right.$ has the positive integer solution
$(a, b)=(13,53),(31,53)$.
Upon verification, when $(a, b)=(13,53)$, it contradicts $74=r<a+b=66$;
$$
\begin{array}{l}
\text { When }(a, b)=(31,53) \text {, } \\
a^{2}+b^{2}=3770=44 \times 84+74,
\end{array}
$$

satisfies the conditions.
In summary, there is only one solution
$(a, b, q, r)=(31,53,44,74)$.
At this time, $a b=31 \times 53=1643$.