jmc

First we change everything to base 10: $$\begin{array}{rl}16A+4B+C+200& =81A+9B+C\Rightarrow \\ 200& =65A+5B.\end{array}$$Notice that the $C$ cancels out on both sides, so $C$ could be any valid digit that works in both base 4 and 9 (0, 1, 2, 3). Now we maximize $A$ with $A=3$ and solve $200=65(3)+5B$ to get $B=1$. If we take any smaller value for $A$, then $B$ will be too large to be a digit. So there is only one value of $A$, one value of $B$, and four possible values for $C$. The sum is $3+1+0+1+2+3=\boxed{10}$.