Sum of those numbers

The only such numbers are $11,12,15,24,36$ , with total $\boxed{98}$ . These are easy to find computationally, but a manual method is as follows: say the number is $10a+b$ , where $1 \le a \le 9$ and $1 \le b \le 9$ (note that neither digit can be zero).

The equation we need these to satisfy is $10a+b=kab$ , where $k$ is a positive integer.

Note that $b=(kb-10)a$ ; so $a$ divides $b$ . We can write $b=ha$ for some positive integer $h$ . Since $a\le 9$ , we have $h \le 9$ as well.

The equation now becomes $10a+ha=kha^2$ , which simplifies (since $a>0$ ) to $10+h=kha$ . Dividing by $h$ , we get $\frac{10}{h}+1=ka$ . So $h$ must be a factor of $10$ , and since it's less than or equal to $9$ , this only leaves the set $\{ 1,2,5 \}$ . It's now a short case analysis to find the above list:

Case $h=1$ : $ka=11$ ; only solution is $a=1,b=1$

Case $h=2$ : $ka=6$ ; solutions $a=1,b=2$ ; $a=2,b=4$ ; $a=3,b=6$

(note we exclude $a=6$ as this would give $b=12$ , which is too large)

Case $h=5$ : $ka=3$ ; only solution is $a=1,b=5$

Incidentally, these numbers appear here in the OEIS.