Note that if $n$ is odd then $d$ is odd and $n+d$ is always even. On the other hand, if $n$ is even then $1$ is a divisor of $n$ and $n+1$ is odd. Therefore it should follow that $n$ is a _winning position_ if and only if $n$ is even. Starting from $1$, the second player has always a winning strategy: replace the current (even) number $n$ with $n+1$.