There are $n$ cards. Max and Lewis play, alternately, the following game
Max starts the game, he removes exactly $1$ card, in each round the current player can remove any quantity of cards, from $1$ card to $t+1$ cards, which $t$ is the number of removed cards by the previous player, and the winner is the player who remove the last card. Determine all the possible values of $n$ such that Max has the winning strategy.
1. **Base Case Analysis:**
   - For \( n = 1 \):
     - Max removes the only card and wins. Hence, Max has a winning strategy.
   - For \( n = 2 \):
     - Max removes 1 card, leaving 1 card for Lewis. Lewis then removes the last card and wins. Hence, Lewis has a winning strategy.
   - For \( n = 3 \):
     - Max removes 1 card, leaving 2 cards for Lewis. Lewis can remove 1 card, leaving 1 card for Max, who then removes the last card and wins. Hence, Max has a winning strategy.
   - For \( n = 4 \):
     - Max removes 1 card, leaving 3 cards for Lewis. Lewis can remove 1 card, leaving 2 cards for Max, who then removes 1 card, leaving 1 card for Lewis, who then removes the last card and wins. Hence, Lewis has a winning strategy.
   - For \( n = 5 \):
     - Max removes 1 card, leaving 4 cards for Lewis. Lewis can remove 1, 2, 3, or 4 cards. In any case, Max can always adjust his move to ensure he removes the last card. Hence, Max has a winning strategy.

2. **Inductive Step:**
   - Assume for \( n = k \), Max has a winning strategy if \( k \equiv 1 \pmod{5} \) or \( k \equiv 4 \pmod{5} \), and Lewis has a winning strategy otherwise.
   - Consider \( n = k + 1 \):
     - If \( k \equiv 0 \pmod{5} \), then \( k + 1 \equiv 1 \pmod{5} \). Max can remove 1 card, leaving \( k \) cards, where \( k \equiv 0 \pmod{5} \). By the inductive hypothesis, Lewis has a winning strategy for \( k \equiv 0 \pmod{5} \), so Max has a winning strategy for \( k + 1 \equiv 1 \pmod{5} \).
     - If \( k \equiv 3 \pmod{5} \), then \( k + 1 \equiv 4 \pmod{5} \). Max can remove 1 card, leaving \( k \) cards, where \( k \equiv 3 \pmod{5} \). By the inductive hypothesis, Lewis has a winning strategy for \( k \equiv 3 \pmod{5} \), so Max has a winning strategy for \( k + 1 \equiv 4 \pmod{5} \).
     - If \( k \equiv 1 \pmod{5} \), then \( k + 1 \equiv 2 \pmod{5} \). Max can remove 1 card, leaving \( k \) cards, where \( k \equiv 1 \pmod{5} \). By the inductive hypothesis, Max has a winning strategy for \( k \equiv 1 \pmod{5} \), so Lewis has a winning strategy for \( k + 1 \equiv 2 \pmod{5} \).
     - If \( k \equiv 2 \pmod{5} \), then \( k + 1 \equiv 3 \pmod{5} \). Max can remove 1 card, leaving \( k \) cards, where \( k \equiv 2 \pmod{5} \). By the inductive hypothesis, Max has a winning strategy for \( k \equiv 2 \pmod{5} \), so Lewis has a winning strategy for \( k + 1 \equiv 3 \pmod{5} \).
     - If \( k \equiv 4 \pmod{5} \), then \( k + 1 \equiv 0 \pmod{5} \). Max can remove 1 card, leaving \( k \) cards, where \( k \equiv 4 \pmod{5} \). By the inductive hypothesis, Max has a winning strategy for \( k \equiv 4 \pmod{5} \), so Lewis has a winning strategy for \( k + 1 \equiv 0 \pmod{5} \).

3. **Conclusion:**
   - By induction, Max has a winning strategy if \( n \equiv 1 \pmod{5} \) or \( n \equiv 4 \pmod{5} \). Otherwise, Lewis has a winning strategy.

The final answer is \( \boxed{ n \equiv 1, 4 \pmod{5} } \)