Let $O$ be the centre of the circle 1) Three points on the circle form an acute triangle if and only if the triangle contains $O$ Proof: follows from that fact that if $ABC$ contains $O$ then $2\angle ABC=\angle AOC$, and the fact that opposite angles of cyclic quadrilaterals sum to $\pi$ 2) The game ends when a string of $n$ or more consecutive points on the circle has been removed Join all remaining points to form an $m$-gon, then triangulate it. If $O$ is in the $m$-gon then it is inside some (acute) triangle. Otherwise $O$ is outside the $m$-gon and all points lie in one half of the plane. This means there is some string of $n$ or more consecutive points that has been removed. 3) The second player always wins. When $n=2$ there are $5$ points, and the second player wins in their first turn by taking a point that is a neighbour of the point that the first player selected. If $n=k>2$ then number the points clockwise $1,2,...,2k+1$. WLOG player one removes point $1$, then player two removes point $k+1$. Notice now that any string of $k-1$ consecutive points must pass over or end at the point $1$ or $k+1$, so we can play the strategy for $2(k-1)+1$ points and insure a win. So, by induction, the second player always wins.

The above solution is wrong! The string $k+2\cdots 2k+1$ has $k$ points and passes through neither $1$ nor $k+1$, so when we delete $1, k+1$ we do not get a reduced game.

Here is a solution in the spirit of the above, but with different execution. Very nice problem Consider the $2n+1$ collections of $n$ contiguous points $A_1, \ldots, A_{2n+1}$ (indices mod $2n+1$). The game proceeds as follows: In a move, one picks an integer $k$ and decrements $S_k, S_{k+1}, \ldots, S_{k+n}$, where $S_k$ is defined as $|A_k|$. A player wins when some $S_i$ becomes zero. Another restriction we impose is that no integer $k$ can be picked twice. This allows for the following transformation: Given $2n+1$ piles of stones, each consisting of $n$ stones, a move consists of picking $n$ consecutive piles that have not been chosen before and removing a stone from each of the piles. The win condition is to fully deplete a condition; equivalently, the first player to make a pile have $1$ stone loses, since the other player now just picks the last block of $n$ piles that contains the pile with $1$ stone left and wins. Thus we play the game where the player to make a pile have exactly $1$ stone loses. Now, we claim that the second player always wins. For a block $P_1\dots P_n$ of stones, we say that the piles $P_0$ and $P_{n+1}$ are the two pivots of the block. Also, note that each pile is the pivot of two blocks. Thus, we may represent the situation as a graph $G$ in which blocks are vertices and piles are edges. Since $(n,2n+1) = 1$, $G$ is a $C_{2n+1}$, and players alternate turns picking vertices. Note that at any point, the vertices chosen form a bunch of disjoint paths. We claim that second player can always ensure that after their turn, each path has even length. We induct on the number of turns; after 0 this is clear. Now once first player moves, one path has odd length, and so second player just picks a vertex so as to extend the odd path (it's fine if this merges with another path, since we just add an even number of edges on to the now-even path). To see why this wins, note that after $2m$ vertices have been chosen, they can be partitioned into $m$ edges. Thus if we call the pivots $Q_1, \ldots, Q_m$, we see that we have removed $m$ stones from each pile but $m-1$ from $Q_1, \ldots, Q_m$. This results, when $m = n-2$, into us having $n-2$ piles with $3$ stones each, and $n+3$ piles of $2$ stones each. It is clear that first player's next move must remove a stone from one of the piles with $2$ stones, so first player must lose. $\blacksquare$