Is the answer all $n \neq 2,3$ ?

Could you please give me some guidance ? Thank you.

Help me please...

Sorry I miss this condition as the last coin just taken away by the opponent in the previous step

The answer is all $n\ge 4$. Clearly $X$ will win for $n=2,3$. We rewrite the game as such: Two players $X,Y$ take turns writing $2$ distinct numbers on a board among $\{1,2,\ldots ,n\}$ with $X$ going first. They cannot write a number written in the last turn by their opponent and each pair of numbers can be written at most twice. We show $Y$ can always win for $n\ge 4$. We pair the pair of numbers, such that whenever $X$ writes a pair $Y$ can write a corresponding pair. For $n=4$, pair as such: $\{(1,2),(3,4)\}, \{(1,3),(2,4)\}, \{(1,4),(2,3)\}$. For $n=5$, $\{(1,2),(3,4)\}, \{(1,3),(2,5)\}, \{(1,4),(3,5)\}, \{(1,5),(2,4)\}, \{(2,3),(4,5)\}.$ For $n=6$, $\{(1,3),(2,5)\}, \{(1,4),(2,6)\}. \{(1,5),(3,6)\}, \{(1,6),(4,5)\}, \{(2,3),(4,6)\}, \{(2,4),(3,5)\}, \{(1,2),(3,4),(5,6)\}$. For the final triplet, $Y$ must ensure that (s)he does not use the same pair until the last matchup to win(e.g. if $X$ writes $(1,2)$, $Y$ writes $(3,4)$, if $X$ writes $(1,2)$ again(may be some time later) $Y$ must write $(5,6)$(not the same pair $Y$ written previously))(note that each pair can be written twice). For $n=7$, $\{(1,3),(4,6)\}, \{(1,4),(5,7)\}. \{(1,5),(4,7)\}, \{(1,6),(2,7)\}, \{(1,7),(3,5)\}, \{(2,3),(4,5)\}, \{(2,4),(6,7)\}, \{(2,5),(3,6)\}, \{(2,6),(3,7)\}, \{(1,2),(3,4),(5,6)\}$. Hence $Y$ wins for $n=4,5,6,7$. We now show if $Y$ can win for $n$, then $Y$ wins for $n+4$. Within $n+1$ to $n+4$ we can pair as such: $\{(n+1,n+2),(n+3,n+4)\}, \{(n+1,n+3),(n+2,n+4)\}, \{(n+1,n+4),(n+2,n+3)\}$. For the others, we pair $\{(n+1,1),(n+2,2)\},\{(n+1,2),(n+2,3)\},\ldots ,\{(n+1,n),(n+2,1)\}$ and $\{(n+3,1),(n+4,2)\},\{(n+3,2),(n+4,3)\},\ldots ,\{(n+3,n),(n+4,1)\}$. Hence $Y$ can win for all $n\ge 4$.

nice solution thank you