jmc

We start by changing the expressions to base 10 in terms of $a$ and $b$. We also know that the two expressions should be equal since they represent the same number. $$\begin{array}{rl}32_a& =23_b\Rightarrow \\ 3\cdot a+2\cdot 1& =2\cdot b+3\cdot 1\Rightarrow \\ 3a+2& =2b+3\Rightarrow \\ 3a& =2b+1.\end{array}$$For the smallest sum $a+b$, we would want the smallest bases $a$ and $b$. $a$ and $b$ must be greater than 3, so we'll let $a=4$ and that means $12=2b+1$ and $b$ is not an integer. Next we try $a=5$ and that means $15=2b+1$ and $b=7$. That means our sum is $5+7=\boxed{12}$. We can check that both expressions work: $32_5=3\cdot 5+2=17$ and $23_7=2\cdot 7+3=17$. Also, it makes sense that $a=5$ works while $a=4$ does not since $3a$ must be odd for $b$ to be an integer ($3a-1=2b$ means $3a$ must be even after subtracting 1), and for $3a$ to be odd, $a$ must also be odd.

Alternatively, we can just try different bases. The smallest possible value for $a$ and $b$ is 4. If we let $b=4$, we'd need a smaller base for $a$ (since we have $3\cdot a\approx 2\cdot b$), which isn't possible. When we let $a=4$, we get $32_4=14$ and try to find a $b$ such that $23_b=14$. That means $2b+3=14$ and $b$ is not an integer. When we let $a=5$, we get $32_5=17$ and try to find a $b$ such that $23_b=17$. If $2b+3=17$, then $b=7$ and we still get $a+b=\boxed{12}$.