2001 AIME I Problem 1

If the number is $10a + b,$ divisibility by the tens digit $a$ forces $b$ to be a multiple of $a$

Write $b = ka.$ Then $10a + b$ divisible by $b$ forces $k = 1,$ $2,$ or $5$

Let the number be $10a + b$ with tens digit $a$ and units digit $b.$ Since $10a + b$ is divisible by $a,$ we need $b$ to be divisible by $a,$ so $b = ka$ for some positive integer $k.$ Since $10a + b$ is divisible by $b,$ we need $10a$ to be divisible by $b,$ that is $10a$ is divisible by $ka,$ so $10$ is divisible by $k.$ Because $b = ka \le 9,$ only $k = 1,$ $2,$ and $5$ are possible.

For $k = 1$ the numbers are $11, 22, \ldots, 99,$ with sum $11 \cdot 45 = 495.$ For $k = 2$ they are $12, 24, 36, 48,$ with sum $120.$ For $k = 5$ the only one is $15.$

The total is $495 + 120 + 15 = 630.$