Find the sum of all positive two-digit integers that are divisible by each of their digits.
Let our number be $10a + b$, $a,b \neq 0$. Then we have two conditions: $10a + b \equiv 10a \equiv 0 \pmod{b}$ and $10a + b \equiv b \pmod{a}$, or $a$ divides into $b$ and $b$ divides into $10a$. Thus $b = a, 2a,$ or $5a$ (note that if $b = 10a$, then $b$ would not be a digit).  

For $b = a$, we have $n = 11a$ for nine possibilities, giving us a sum of $11 \cdot \frac {9(10)}{2} = 495$.
For $b = 2a$, we have $n = 12a$ for four possibilities (the higher ones give $b > 9$), giving us a sum of $12 \cdot \frac {4(5)}{2} = 120$.
For $b = 5a$, we have $n = 15a$ for one possibility (again, higher ones give $b > 9$), giving us a sum of $15$.
If we ignore the case $b = 0$ as we have been doing so far, then the sum is $495 + 120 + 15 = \boxed{630}$.