Solution:

Let the number be $10a + b$ with tens digit $a$ and units digit $b.$ Since $a \mid 10a + b,$ we need $a \mid b,$ so $b = ka$ for some positive integer $k.$ Since $b \mid 10a + b,$ we need $b \mid 10a,$ that is $ka \mid 10a,$ so $k \mid 10.$ 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 $495 + 120 + 15 = 630.$