jmc

The square connected both to 1 and 2 cannot be the same as either of them, so must be 3. $$\begin{array}{ccc}1& 3& \\ & 2& A\\ & & B\end{array}$$ The last square in the top row cannot be either 1 or 3, so it must be 2. $$\begin{array}{ccc}1& 3& 2\\ & 2& A\\ & & B\end{array}$$ The other two squares in the rightmost column with A and B cannot be two, so they must be 1 and 3 and therefore have a sum of $1+3=\boxed{4}$.