I recently came across the "Taxman Game" the rules are in the link, but I'll repeat them here:
We start with a pile of integers, from 1 to some number that you choose [$n$]. You take one, and I get all the others that divide it evenly. We repeat until there's nothing left. BUT (big but!) I have to get something on each turn. So when none of the numbers in the list have any divisors that have not been taken, I take them all. Your score is the sum of all the numbers you took, while my score is the sum of all the numbers I took.
The 'you' and 'I' is a bit confusing, there is only one player here - the 'you' - as the 'taxman' ('I') side is entirely deterministic.
The question I have is this: for what $n$ is the game winnable? I think that it is winnable for all $n>3$ but I'm not sure how to go about proving this. At $n=3$, the best strategy is a draw (pick 3, taxman get 1 and 2), while for $n=1$, the player has no winning move so the lowest bound on winnable $n$ must be $n>3$ but is there a higher $n$ that is unwinnable?
I'd also be interested in whether there is a 'best' strategy guaranteed to give a win where it is possible and the largest possible winning margin when doing so.