Find all the numbers

Let $t$ satisfy $(10a + b)t = 100a + b$ , with $a,b \neq 0$ and $b<10$ . Clearly $t < 10$ .

Then $b(t-1) = 10a(10-t)$ . Thus $b(t-1)$ is a multiple of $10$ . Given that $b<10$ and $t-1<9$ , and $5$ must be a prime factor of $b(t-1)$ , either $b=5$ or $t-1=5$ , i.e. $t=6$ .

If $b = 5$ , then $t$ is an odd number between $2$ and $10$ . But $t=3$ and $t=5$ are too small, since $(10-t)*10$ divides $b(t-1)$ .

If we try $t=7$ , then $b(t-1) = 30 = 3*10*a$ , which works with $a=1$ . We can verify $15$ is a solution, since $105=7*15$ .

If we try $t=9$ , then $b(t-1) = 40 = 1*10*a$ , which works with $a=4$ . We can verify $45$ is a solution, since $405 = 9*45$ .

The other option uses $t-1 = 5$ , i..e. $t=6$ . Then $5b = 10a*4$ , i.e. $b = 8a$ . Since $b$ is a single digit number, the only solution is $a=1, b=8$ . We can verify $18$ is a solution since $108=6*18$ .

We have exhausted the possible values of $a$ , $b$ , and $t$ . Thus the only solutions are $15$ , $45$ , and $18$ , which sum to $78$ .

We are looking for positive integers $a,b$ such that $1 \le b \le 9$ and $10a + b$ divides $100a + b$ . But then $10a + b$ must divide $10(10a+b) - (100a+b) = 9b$ , and it is easy to check that the only pairs of integers $(a,b)$ with these properties are $(1,5)$ , $(4,5)$ and $(1,8)$ . Thus the answer is $15 + 45 + 18 = \boxed{78}$ .