PARTIAL ANSWER

Here is a winning strategy for $n=100$

Take $97,$ even numbers $100-68, 64, 60,56,52,99,93,87,81,75,69,63,95,85,91$. This gives a total of $2595$ coins, over half of the $5050$ in play. As Daniel Mathias says, you start with the largest prime, then take the even numbers down to $\frac {2n}3$ with half the number the witness that you can take it, then take multiples of $4$ from $\frac {2n}3$ to $\frac n2$. You skip the numbers equivalent to $2 \pmod 4$ in the bottom range because you want the number $\frac 32$ times as large in the next step. For example, we skip $62$ because it would use $31$ and we want $93$. We don't skip $60$ because $90=2 \cdot 45$ will still be available. We then take the available multiples of $3$ down to $63$, and finally $95,85,91$. I don't know if this approach will always win.

PARTIAL ANSWER

Here is a winning strategy for $n=100$

Take $97,$ even numbers $100-68, 64, 60,56,52,99,93,87,81,75,69,63,95,85,91$. This gives a total of $2595$ coins, over half of the $5050$ in play. As Daniel Mathias says, you start with the largest prime, then take the even numbers down to $\frac {2n}3$ with half the number the witness that you can take it, then take multiples of $4$ from $\frac {2n}3$ to $\frac n2$. You skip the numbers equivalent to $2 \pmod 4$ in the bottom range because you want the number $\frac 32$ times as large in the next step. For example, we skip $62$ because it would use $31$ and we want $93$. We don't skip $60$ because $90=2 \cdot 45$ will still be available. We then take the available multiples of $3$ down to $63$, and finally $95,85,91$. I don't know if this approach will always win. There are more points available from small numbers before you do the series I listed. I don't know if they are ever necessary to win.