Let's find a number written in the decimal system that is 9 times larger than the number we get by reading its digits in reverse order.
Let the sought number be denoted by $K$, and its reverse by $F$. Ignoring the trivial solution $K=0$, the first digit of $K$ cannot be 0, so neither the last digit of $F$ nor the last digit of $K=9F$ can be 0. Let the number of digits in $K$ be denoted by $n$, then

$$
10^{n-1} \leq K=9 F \leq 10^{n}-1
$$

Therefore, $F$ cannot be larger than the $n$-digit number where every digit is 1. Thus, the first digit of $F$ is 1 and the first digit of $K$ is 9, i.e., $K=9 \ldots 1, F=1 \ldots 9$. Let's examine what the second digit of $F$ can be.

Since $F \leq 11 \ldots 1$, the second digit of $F$ can only be 1 or 0. If $F=11 \ldots 9, K=9 \ldots 11$, then the second digit of $K$ could only be 7, but $K=97 \ldots 11<9 F$. If $F=10 \ldots 9, K=9 \ldots 01$, then the second digit of $K$ can only be 8: $F=10 \ldots 89, K=98 \ldots 01$. It is easy to see that for the numbers $K=9801, F=1089$, the condition $K=9 F$ is satisfied, so $K$ can be 9801.

Remarks. 1. To find further solutions, we need to examine for which $k$ the following equation holds:

$$
\overline{98 k 01}=9 \cdot \overline{10 f 89}
$$

where $f$ is the reverse of $k$; the overline denotes the concatenation of digits. From this, with some calculation, we get the condition

$$
9 f-k=8\left(10^{n-4}-1\right)
$$

which is always satisfied by numbers consisting entirely of the digit 9. This means that in the form $K=98 \ldots 01$, we can place any number of 9s in the place of the dots and always get a valid number.

2. It can also be shown that if $K$ and $L$ are valid numbers, then $\overline{K L K}$ is also a valid number, for example, from the numbers $K=9801, L=98901$, we get the number 9801989019801, so in the form $K=98 \ldots 01$, other digits can also stand in the place of the dots, not just all 9s.