Let the number of marbles George has be $x$, and so the number of marbles Jane has is $2x$. Therefore, the total number of non-chipped marbles is $3x\equiv0 \ (\text{mod }3)$. However, the total number of marbles is $18+19+21+23+25+34=140\equiv2 \ (\text{mod }3)$. Therefore, we need a number of chipped marbles $\equiv2 \ (\text{mod }3)$ to get a number of non-chipped marbles $\equiv0 \ (\text{mod }3)$.
$18\equiv0 \ (\text{mod }3)$
$19\equiv1 \ (\text{mod }3)$
$21\equiv0 \ (\text{mod }3)$
$23\equiv2 \ (\text{mod }3)$
$25\equiv1 \ (\text{mod }3)$
$34\equiv1 \ (\text{mod }3)$
Since $23$ is the only one $\equiv2 \ (\text{mod }3)$, it is the only possible number of chipped marbles, $\boxed{\text{D}}$.
To check, we can see that if Jane takes the $19, 25,$ and $34$ marble bags and George takes the $18$ and $21$ marble bags, this satisfies the conditions.