First we change everything to base 10: \begin{align*} 16A+4B+C+200&=81A+9B+C\quad\Rightarrow\\ 200&=65A+5B.\end{align*} 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} $.