9. Find such four-digit numbers, each of which is divisible by 11, and the sum of the digits of each of them is 11.
9. Let the desired number be denoted as: $1000 x+100 y+10 z+u$, then $x+y+z+u=11$. According to the divisibility rule for 11 (if the sum of the digits of a number at odd positions equals the sum of the digits at even positions or the difference between these sums is divisible by 11, then the number is divisible by 11), we can write $\frac{(x+z)-(y+u)}{11}=k$.

Since $-11 \leqslant(x+z)-(y+u) \leqslant 11$, $k$ can only be $0, -1$, or +1

$$
\begin{aligned}
& +\begin{array}{l}
(x+z)+(y+u)=11 \\
(x+z)-(y+u)=11 k \\
\hline 2(x+z)=11(k+1)
\end{array}
\end{aligned}
$$

Let $k=-1$, then $2(x+z)=0$, i.e., $x+z=0$. This is impossible.

Let $k=1$, then $x+z=11, y+u=0$, i.e., $y=0, u=0$.

If we take $k=0$, then $x+z=\frac{11}{2}$, which is impossible.

Answer. The problem has eight solutions: 2090, $3080,4070,5060,6050,7040,8030,9020$.