jmc

We can rewrite $AA{A}_4$ and $33_b$ to get $$\begin{array}{rl}16A+4A+A& =3b+3\Rightarrow \\ 21A& =3b+3.\end{array}$$The smallest possible value for $A$ is 1, which gives us $21=3b+3$ and $b=6$. So the smallest sum $A+b=\boxed{7}$. While there are other values for $A$ and $b$ that work, increasing $A$ will increase $b$, resulting in a larger sum.