552. Find all four-digit numbers that are 9 times greater than their reversals.
$\triangle$ Let the desired number be denoted as $\overline{a b c d}$. Then

$$
\overline{a b c d}=9 \cdot \overline{d c b a}
$$

Two digits, $d$ and $a$, can be found immediately. Since the product $9 \cdot \overline{d c b a}$ is a four-digit number, the digit $d$ can only be 1. Since the product $9a$ ends with the digit $d=1$ (consider the last digit of each four-digit number), then $a=9$.

The number $\overline{a b c d}=\overline{9 b c 1}$ is divisible by 9, so the sum of its digits is divisible by 9:

$$
(10+b+c): 9
$$

Thus, $b+c=8$ or $b+c=17$.

Now, let's use the statement from problem 550, considering that the number of digits in $\overline{a b c d}$ is even: the sum

$$
\overline{a b c d}+\overline{d c b a}
$$

must be divisible by 11. Transform it:

$$
\overline{a b c d}+\overline{d c b a}=9 \cdot \overline{d c b a}+\overline{d c b a}=10 \cdot \overline{d c b a}=10 \cdot \overline{1 c b 9}
$$

Then the number $\overline{1 c b 9}$ is divisible by 11. By the divisibility rule for 11, the sum

$$
1+b-c-9=b-c-8
$$

is divisible by 11. Therefore, $b-c=8$ or $b-c=-3$.

Compare the values of $b+c$ with the values of $b-c$. Since the sum and difference of two integers have the same parity (see the statement of problem 93 in § 4), out of the four possible cases here, it is sufficient to consider two.

1) The system of equations

$$
\left\{\begin{array}{l}
b+c=8 \\
b-c=8
\end{array}\right.
$$

has the solution $b=8, c=0$.

2) The system of equations

$$
\left\{\begin{array}{l}
b+c=17 \\
b-c=-3
\end{array}\right.
$$

has the solution $b=7, c=10$, but it is not valid.

Answer: 9801.