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 $a b=10 a+b$ be a two-digit number with digits $a$ and $b$ that is divisible by the sum $a+b$ of its digits. Note that, being a two-digit number, necessarily $a \neq 0$ and that, being divisible by the sum of its digits, the difference $(10 a+b)-(a+b)=9 a$ is also divisible by $a+b$ (prove this). Thus, it suffices to assign the values $1,2,3,4,5,6,7,8$ and 9 to $a$ and calculate the values of $b$ for which $a+b$ divides $9 a$. The result appears in the table.

| $a$ | $9 a$ | $b$ |
| :---: | :---: | :---: |
| 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 |

Thus, the numbers that satisfy the property are

$10,12,18,20,21,24,27,30,36,40,42,45,48,50,54,60,63,70,72,80,81,84$ and 90, that is, there are 23 "curious" numbers.