The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think). If I have a secret number and you try to guess a larger one, what are your odds of success? is there a way to make them better (if you have $99$ friends). You have only finitely many ways to fail and infinitely many to succeed, but we can't really assign a probability. That is also behind this puzzle

Going back to the given problem, suppose that each box has a guard. At a certain time each guard sees the contents of every other box then guesses the contents of their own. Under AoC with an agreed set of equivalence class representatives and the familiar protocol we know that all but finitely many will be correct. That is strange but we are used to this consequence of the axiom of choice. If you have full knowledge of all this and are required to become the guard for a box of your choice (after looking into as many other boxes as you wish)  can we say that you are **certain** to make a correct guess? Not really.

The situation in the question above seems clearer to me if we stipulate ahead of time that all but finitely many boxes contain $0$  (no assumption , other than finiteness on which the exceptions are).  If there are $100$ potential guards who are friends, can they be sure that each can pick a box and guess $0$ (what else can they do?) with at least $99$ correct? yes, as before split the boxes into $100$ subsequences, person $k$ looks at all the boxes in every subsequence other than his own then chooses a box further than the last non-zero box in every other subsequence.

If you are one member of this set of $100$ friends are your odds of being correct (when you guess $0$) improved $100$ fold, or at all? Not really.

Essentially $100$ people are assigned hidden integers and must name an integer greater than their assigned integer (so only finitely ways to fail and infinitely many to succeed!). If each person can see all the other integers then only one will fail. 


In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) there is no loss in saying that the chosen sequence is in a certain coset (so why not the coset of the zero sequence)?