First we consider integers that have $ 2 $ digits in base $ 2 $ and $ 1 $ digit in base $ 3 $. Such an integer must be greater than or equal to $ 10_2 = 2 $, but strictly less than $ 10_3 = 3 $. The only such integer is $ 2 $. Next we consider integers that have $ 4 $ digits in base $ 2 $ and $ 2 $ digits in base $ 3 $. Such an integer must be greater than or equal to $ 1000_2 = 2^3 $, but strictly less than $ 100_3 = 3^2 $. The only such integer is $ 8 $. Next we consider integers that have $ 6 $ digits in base $ 2 $ and $ 3 $ digits in base $ 3 $. Such an integer must be greater than or equal to $ 100000_2 = 2^5 $, but strictly less than $ 1000_3 = 3^3 $. There are no such integers, because $ 2^5 > 3^3 $. If we continue in this fashion, we may come to suspect that there are no more solutions of any length. Let us prove this. If an integer $ N $ has $ 2d $ digits in base $ 2 $, then $ N\ge 2^{2d-1} $. But if $ N $ has only $ d $ digits in base $ 3 $, then $ N<3^d $. A mutual solution is possible only if $$2^{2d-1}<3^d.$$We can rearrange this inequality as $$\left(\frac 43\right)^d < 2.$$By inspection, this inequality is valid for $ d=1,2 $ but invalid for $ d=3 $, and also invalid for any larger $ d $ since the left side increases as $ d $ increases. This shows that there are no solutions $ N $ beyond those we found already: $ 2 $ and $ 8 $, whose sum is $ \boxed{10} $.