jmc

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 $${\left(\frac{4}{3}\right)}^d<2.$$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 $\boxed{10}$.