Let's consider a four-digit natural number with the following property: if we swap its first two-digit number with the second, we get a four-digit number that is 99 less.

How many such numbers are there in total, and how many of them are divisible by 9? (K. Pazourek)
Let the original four-digit number be $\overline{a b c d}=1000 a+100 b+10 c+d$. According to the problem, $\overline{a b c d}=\overline{c d a b}+99$, where $b$ and $d$ are integers from 0 to 9, $a$ and $c$ are integers from 1 to 9, and $a \geq c$. By expanding and rearranging the above equation, we get:

$$
\begin{aligned}
1000 a+100 b+10 c+d & =1000 c+100 d+10 a+b+99 \\
990 a-990 c & =99 d-99 b+99, \\
10(a-c) & =d-b+1 .
\end{aligned}
$$

From the above constraints, the number $d-b+1$ on the right side can take values from -8 to 10. From the equation, it follows that this number must be a multiple of 10, i.e., either 10 or 0. We will consider both possibilities separately:
a) $10(a-c)=d-b+1=10$, meaning $a=c+1, d=9$ and $b=0$. There are a total of eight numbers of this form (for $c$ from 1 to 8):

$$
2019,3029,4039,5049,6059,7069,8079,9089 .
$$

Among them, only the number 5049 is divisible by nine (the sum of the digits $2 c+1+9$ is divisible by nine for $c=4$).
b) $10(a-c)=d-b+1=0$, meaning $a=c$ and $b=d+1$. In this case, we can substitute $c$ from 1 to 9 and $d$ from 0 to 8, i.e., a total of $9 \cdot 9=81$ possibilities:

$$
\begin{array}{ccc}
1110, \ldots, & 1918, \\
\vdots & & \vdots \\
9190, \ldots, & 9998 .
\end{array}
$$

The sum of the digits of numbers of this form is $2(c+d)+1$, and this sum is divisible by nine within our constraints either for $c+d=4$, or for $c+d=13$. In the first case, we get four numbers:

$$
1413,2322,3231,4140 .
$$

In the second case, we get five numbers:

$$
5958,6867,7776,8685,9594 \text {. }
$$

In total, there are $8+81=89$ numbers that satisfy the first part of the problem. Among these numbers, $1+4+5=10$ are divisible by nine.

Evaluation. 2 points for preparatory manipulations and observations; 2 points for the result; 2 points for completeness and quality of the commentary.

Note. The problem can be viewed as an alphametic:

$$
\begin{array}{r}
c d a b \\
+\quad 99 \\
\hline a b c d
\end{array}
$$

Some of the previous considerations may be more illustrative in this approach.