Question:
What is the sum of all positive integers that have twice as many digits when written in base $2$ as they have when written in base $3$? Express your answer in base $10$.

Answer:
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 $\boxed{10}$.