Main claim: Let L be and integer, then for an L-tuple S of positive integers, define f(S) to be the sum of their reciprocals. Then the fractional part of f(S) is bounded above by a number less than 1 that solely depends on L. Proof: Induction. Just note that if the fractional part can get infinitely close to 1, that means that the smallest reciprocal in the sum must be infinitely small, but by removing it, we get an L-1 tuple of reciprocals who's sums fractional part is infinitely close to 1. Contradiction. Alice strategy: Now, all that she has to do is choose very very big numbers, so that their reciprocals are very small. By doing so, the only way that the number would become integer after some turn, the fractional part of the sum of all numbers, except the one that Alice selected last, would have to be very very close to one, however, by the main claim, Alice can choose her numbers in such a way, so that would be impossible.