N2 (2-1, Bulgaria) Find all three-digit numbers divisible by 11, such that the quotient equals the sum of the squares of the digits of the three-digit number.
Let the quotient obtained by dividing this three-digit number by 11 be $\overline{a b}$. If $a+b \geqslant 10$, then the original number is
$$
100(a+1)+10(a+b-10)+b ;
$$

If $a+b>10$, we have
$$
\begin{aligned}
& (a+1)^{2}+(a+b-10)^{2}+b^{2} \\
\geqslant & (a+1)^{2}+1+(11-a)^{2} \\
= & 2\left(a-\frac{15}{2}\right)^{2}+10 a+\frac{21}{2} \\
> & 10 a+b,
\end{aligned}
$$

which contradicts (1). Therefore, we must have $a+b=10$. Substituting $b=10-a$ into (1), we get
$$
2 a^{2}-27 a+91=0 .
$$

Solving this, we find $a=7$, and thus $b=3$.
(2) If $a+b<5$, we have
$$
\begin{aligned}
& a^{2}+(a+b)^{2}+b^{2} \\
= & 2(a+b) a+2 b^{2} \\
> & 10 a+b,
\end{aligned}
$$

which contradicts (2). Therefore, $a+b \leqslant 5$. Since the left side of (2) is even and $10 a$ is also even, $b$ must be even, so $b$ can only be $0, 2, 4$.

When $b=0$, $a=5$; when $b=2$, $a$ is not an integer; when $b=4$, $a$ is also not an integer.
In summary, $\overline{a b}=73$ or 50, and the original two-digit number is 803, 550.