jmc

Let us represent a two-digit integer by $ab$, where $a$ is the tens digit and $b$ is the units digit. Then the value of the number is $10a+b$, the sum of the digits is $a+b$, and the product of the digits is $a\cdot b$. We are given that $a+b\mid 10a+b$ and $ab\mid 10a+b$. We know neither $a$ nor $b$ is zero since nothing is divisible by zero. We work with the equation $a+b\mid 10a+b$. We also know that $a+b\mid a+b$, so $a+b$ must divide the difference, which is $10a+b-a-b=9a$. So we have $a+b\mid 9a$, or $k(a+b)=9a$ for some integer $k$. Solving this equation gives $kb=(9-k)a$, or $\frac{b}{a}=\frac{9-k}{k}$. Since $a$ and $b$ are both positive, we must have $0<k\le 9$, so the possible values of $\frac{b}{a}$ are $\frac{1}{8},\frac{2}{7},\frac{3}{6},\frac{4}{5},\frac{5}{4},\frac{6}{3},\frac{7}{2},\frac{8}{1}$. For each of these possibilities except $\frac{3}{6}$ and $\frac{6}{3}$, the fraction does not reduce, and thus the only values of $a$ and $b$ which will satisfy them are $a$ as the number in the denominator and $b$ as the number in the numerator. There is no pair of larger $(a,b)$ with the same ratio or else either $a$ or $b$ wouldn't be a single digit, and there is no pair of smaller $(a,b)$ since the fractions do not reduce. For these cases, we check to see whether $ab\mid 10a+b$:

\begin{array}{c|c|c|c} $(a,b)%%DISP_0%%amp;$ab%%DISP_0%%amp;$10a+b%%DISP_0%%amp;Will it divide?\\ \hline $(1,8)%%DISP_0%%amp;$8%%DISP_0%%amp;$18%%DISP_0%%amp;No\\ $(2,7)%%DISP_0%%amp;$14%%DISP_0%%amp;$27%%DISP_0%%amp;No\\ $(4,5)%%DISP_0%%amp;$20%%DISP_0%%amp;$45%%DISP_0%%amp;No\\ $(5,4)%%DISP_0%%amp;$20%%DISP_0%%amp;$54%%DISP_0%%amp;No\\ $(7,2)%%DISP_0%%amp;$14%%DISP_0%%amp;$72%%DISP_0%%amp;No\\ $(8,1)%%DISP_0%%amp;$8%%DISP_0%%amp;$81%%DISP_0%%amp;No \end{array}

The only cases which remain are those for which $\frac{b}{a}=\frac{6}{3}=2$, or $\frac{b}{a}=\frac{3}{6}=\frac{1}{2}$. So we have $b=2a$ or $a=2b$.

If $a=2b$, we must check whether $ab\mid 10a+b$. Substituting, we must find $b$ such that $2b^2\mid 10(2b)+b$, or $2b^2\mid 21b$. This means $21b=m(2b^2)$ for some integer $m$, or (since $b\ne 0$) $21=2mb$. But the right side is even and $21$ is odd, so there are no $b$ which satisfy this and thus no numbers with $a=2b$.

If $b=2a$, again we substitute to find $2a^2\mid 10a+2a$, or $2a^2\mid 12a$. This means $12a=n(2a^2)$ for some integer $n$, or $6=na$, so $a$ must be a divisor of $6$. Thus $a$ can be $1,2,3$, or $6$. The corresponding values of $b$ are $2,4,6$ and $12$. But $b\le 9$, so the pair $(6,12)$ must be thrown out and we have three possible pairs for $(a,b)$: $(1,2)$, $(2,4)$, and $(3,6)$. These correspond to the numbers $12,24$, and $36$, and the sum is $12+24+36=\boxed{72}$.