The number 81 has the following property: it is divisible by the sum of its digits $8+1=9$. How many two-digit numbers satisfy this property?
Let $ab$ be such a number. By hypothesis, $ab = 10a + b$ is divisible by $a + b$. Therefore, the difference $(10a + b) - (a + b) = 9a$ is also divisible by $a + b$. Moreover, we know that $10a + b$ is divisible by $a + b$ if and only if $(10a + b) - (a + b) = 9a$ is divisible by $a + b$ (prove this).

Before proceeding with the solution, note that since $ab$ is a two-digit number, $a \neq 0$. Now, it suffices to assign values to $a$ and calculate the values of $b$ for which $a + b$ divides $9a$. The result is shown in the following table.

| $a$ | $9a$ | $b$ |
| :---: | :---: | :---: |
| 1 | 9 | $0,2,8$ |
| 2 | 18 | $0,1,4,7$ |
| 3 | 27 | 0,6 |
| 4 | 36 | $0,2,5,8$ |
| 5 | 45 | 0,4 |
| 6 | 54 | 0,3 |
| 7 | 63 | 0,2 |
| 8 | 72 | $0,1,4$ |
| 9 | 81 | 0 |

Therefore, the numbers that satisfy the property are:

$10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90$

In other words, there are 23 numbers that meet the required conditions.