(Communicated by Kai Wang) For all $n$, Bob has a winning strategy. Note that we can interpret the game play as building a permutation of $\{1,\dots,n\}$, and the number of times an integer $k$ is chosen on the $k$-th turn is exactly the number of fixed points of this permutation. For $n$ even, Bob selects the goal "even". Divide $\{1,\dots,n\}$ into the pairs $\{1,2\},\{3,4\},\dots$; each time Alice chooses an integer, Bob follows suit with the other integer in the same pair. For each pair $\{2k-1,2k\}$, we see that $2k-1$ is a fixed point if and only if $2k$ is, so the number of fixed points is even. For $n$ odd, Bob selects the goal "odd". On the first turn, if Alice chooses 1 or 2, then Bob chooses the other one to transpose into the strategy for $n-2$ (with no moves made). We may thus assume hereafter that Alice's first move is some $k > 2$, which Bob counters with 2; at this point there is exactly one fixed point. Thereafter, as long as Alice chooses $j$ on the $j$-th turn (for $j \geq 3$ odd), either $j+1 < k$, in which case Bob can choose $j+1$ to keep the number of fixed points odd; or $j+1=k$, in which case $k$ is even and Bob can choose 1 to transpose into the strategy for $n-k$ (with no moves made). Otherwise, at some odd turn $j$, Alice does not choose $j$. At this point, the number of fixed points is odd, and on each subsequent turn Bob can ensure that neither his own move nor Alice's next move does not create a fixed point: on any turn $j$ for Bob, if $j+1$ is available Bob chooses it; otherwise, Bob has at least two choices available, so he can choose a value other than $j$.