Yvan and Zoé are playing the following game. Let $n \in \mathbb{N}$. The integers from 1 to $n$ are written on $n$ cards arranged in order. Yvan removes one. Zoé then removes 2 consecutive ones. Next, Yvan removes 3 consecutive ones, and Zoé finishes by removing 4 consecutive ones.

What is the smallest value of $n$ for which Zoé can ensure she can play both of her turns?
(solved by Erik Desurmont)

For $n=1$ or 2, once a card is removed, Zoé can no longer play. For $n=3$, Zoé cannot play if Yvan removes the second card. For $4 \leqslant n \leqslant 13$, Yvan removes the fourth card. Two cases arise: either Zoé removes cards with a number greater than 4: in this case, there are at most 3 sequences of consecutive cards. Zoé cannot play in the first, and the other two sequences contain at most $9-2=7$ cards. In particular, the smallest sequence contains at most 3 cards, so there is only one possibility for Zoé to play. Either Yvan cannot play, or he takes three cards from the middle of the largest sequence, in which case Zoé cannot play (there will be at most 4 cards left in the sequence, but if there are that many, they will be split into 2, so Zoé cannot play.

If Zoé removes cards that are lower, she cannot remove 4 lower cards on the next turn. Yvan then removes 3 cards if possible from the middle of the $n-4 \leqslant 9$ largest cards, so that there are at most 3 consecutive cards left, which is possible because $3+3+3 \geqslant n-4$. Thus, Zoé cannot play.

Let's show that for $n=14$, Zoé can play two turns. If Yvan takes the first or second card, Zoé takes the next two cards, leaving at least 10 consecutive cards. In particular, by removing 3 cards, 7 cards are left divided into two sequences, so one sequence must contain at least 4 cards, and Zoé can play. If Yvan takes the third or fourth card, Zoé can take the first two cards, leaving 10 consecutive cards, and Zoé can ensure she plays a second turn in the same way. If Yvan takes the fifth, sixth, or seventh card, Zoé takes the next two cards. There will then be $4, 5, 6$ cards on the left and $7, 6, 5$ cards on the right consecutively, and Zoé can play her second move by removing 4 cards from the consecutive sequence in which Yvan did not draw. Since the game is symmetric, the case where Yvan draws the $i$-th card with $i \geqslant 8$ is the same as if Yvan had drawn the $15-i$-th card and $15-i \leqslant 7$. Thus, in all cases, Zoé can play a second turn.

In conclusion, the smallest $n$ for which Zoé is guaranteed to be able to play two turns is $n=14$.