4. 188 Find all three-digit numbers that satisfy the following condition: the quotient when divided by 11 equals the sum of the squares of its digits.
[Solution] Let the required three-digit number be
$$n=100 a+10 b+c$$

where $b, c \in\{0,1,2,3, \cdots, 9\}, a \in\{1,2, \cdots, 9\}$.
The necessary and sufficient condition for $n$ to be divisible by 11 is that $a-b+c$ is divisible by 11. Since
$$-8 \leqslant a-b+c \leqslant 18$$

Thus, only $\square$
$$\begin{array}{r}
a-b+c=0 \\
a-b+c=11
\end{array}$$

Also, by the problem, $100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right)$.
We discuss in two cases:
Case 1 Solve $\left\{\begin{array}{l}a-b+c=0, \\ 100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right)\end{array}\right.$.
From (1) we get
$$\begin{array}{c}
b=a+c \\
00 a+10(a+c)+c=11\left[a^{2}+(a+\right. \\
11(10 a+c)=11\left(2 a^{2}+2 a c+2 c^{2}\right) \\
10 a+c=2 a^{2}+2 a c+2 c^{2} \\
2 a^{2}+2(c-5) a+2 c^{2}-c=0
\end{array}$$

Substituting into (3), we get $100 a+10(a+c)+c=11\left[a^{2}+(a+c)^{2}+c^{2}\right]$,

Considering the discriminant $\Delta=4(c-5)^{2}-8\left(2 c^{2}-c\right) \geqslant 0$,
we get
$$\frac{-\sqrt{91}-4}{3} \leqslant c \leqslant \frac{\sqrt{91}-4}{3}$$

Since $c$ is a non-negative number, therefore
$$c=0 \text { or } c=1 \text {. }$$

But from $\square$
$$10 a+c=2 a^{2}+2 a c+2 c^{2},$$

we know that $c$ must be even, so only $c=0$.
Substituting $c=0$ into (4), we get
$$2 a^{2}-10 a=0$$

But $a \neq 0$, so we get $a=5$.
Also, from $b=a+c \quad$ we get $b=5$
Thus, we obtain the three-digit number 550.
Case 2 Solve $\left\{\begin{array}{l}a-b+c=11, \\ 100 a+10 b+c=11\left(a^{2}+b^{2}+c^{2}\right) .\end{array}\right.$
From (2) we get $b=a+c-11$,
Substituting into (3), we get
$$100 a+10(a+c-11)+c=11\left[a^{2}+(a+c-11)^{2}+c^{2}\right],$$

Simplifying and rearranging, we get
$$2 a^{2}+2(c-16) a+2 c^{2}+131-23 c=0 .$$

Thus, the discriminant $\Delta=4(c-16)^{2}-8\left(2 c^{2}+131-23 c\right) \geqslant 0$, i.e., $\square$
$$-3 c^{2}+14 c-6 \geqslant 0,$$

Hence,
$$\frac{7-\sqrt{31}}{3} \leqslant c \leqslant \frac{7+\sqrt{31}}{3} \text {. }$$

Since $c$ is a non-negative integer, therefore
$$c=1, c=2, c=3, c=4 \text {. }$$

From (6), we can see that $c$ must be odd, so $c=1$ or $c=3$.
Substituting $c=1$ into (6), we get $2 a^{2}-30 a+110=0$,
i.e., $\square$
$$a^{2}-15 a+55=0 .$$

Solving, we get $a=\frac{15 \pm \sqrt{5}}{2}$, which contradicts that $a$ is a positive integer.
Substituting $c=3$ into (6), we get $a^{2}-13 a+40=0$,
Solving, we get $a_{1}=8$ or $a_{2}=5$.
From $b=a+c-11$ we get $b_{1}=0$ or $b_{2}=-3$ (not valid, discard).
Thus, the required three-digit number is 803.
In summary, there are two three-digit numbers that satisfy the conditions: 550 and 803.