Example 5, find a four-digit number that can be divided by 11, and the quotient obtained is 10 times the sum of the digits of this four-digit number.
Let $\overline{a b c d}=110(a+b+c+d)$, then we have
$$
890 a=10 b+100 c+109 d \text {. }
$$

Also, since 11 divides $\overline{a b c d}$, we have $a-b+c-d= \pm 11$ or $a-b+c-d=0$.
If $a-b+c-d=0$,

then from $\left\{\begin{array}{l}a+c=b+d, \\ 890 a=10 b+100 c+109 d\end{array}\right.$
(2) $-10 \times(1)$ we get
$880 a=110 c+99 d \leqslant(110+99) \times 9$
$$
=1881 \text {. }
$$

If $a=1$, substituting into (3) gives $80=10 c+9 d$,
$$
\therefore d=0, c=8, b=9 \text {; }
$$

If $a=2$, substituting into (3) gives $1760=110 c+99 d$
$$
\therefore d=0, c=16 \text { (not valid). }
$$

Therefore, the required four-digit number is 1980.
If $a-b+c-d=11$,
then from $\left\{\begin{array}{l}a+c=11+b+d, \\ 890 a=10 b+100 c+109 d \text { (5) }\end{array}\right.$
(5) $-10 \times$ (4) we get
$110 c=880 a-99 d+110 \geqslant 990$. (6)
$\therefore c \geqslant 9$, i.e., $c=9$. Substituting into (6) gives $a=1$, $d=0, b=-1$ (not valid).
If $a-b+c-d=-11$, similarly, we get no solution.
Therefore, the only solution is 1980.
This problem mainly uses the method of bounding to determine the range of unknowns, which is applicable in a wide range of scenarios.