To solve the problem, let's understand the sequence and the rules of the game.

Initially, all players start with $2024$ pebbles. Each player's turn involves picking a unique number $n$ such that no player has previously picked, taking $2n$ pebbles from their pile, and distributing those pebbles (i.e., $n$ to each of the other two players).

### Game Analysis:

1. Game Dynamics:
- If a player chooses an integer $n$, they lose $2n$ pebbles, and each of the other two players gains $n$ pebbles.
- The sum of pebbles among the three players remains constant at $3 \times 2024 = 6072$.

2. End Condition:
- A player loses the game if on their turn they cannot choose a higher number $n$ such that $2n$ pebbles can be taken from their pile (i.e., their pebbles are fewer than $2n$).

3. Pebbles Allocation:
- Let the player's pebbles count after various rounds be $a_i, b_i, c_i$, for Archie, Billie, and Charlie after round $i$ respectively.
- On Archie's turn in round 1: $a_1 = 2024 - 2n_1$, $b_1 = 2024 + n_1$, $c_1 = 2024 + n_1$.

4. Strategy and Analysis:
- As each progressive round occurs, the number $n$ chosen must always be larger than the previous $n$. This means the depletion in pebbles accelerates.
- The cumulative number of pebbles each player loses over the turns increases due to the growing $n$.

### Strategy to Avoid Loss:

- Charlie's Position:
- Since Charlie plays last in each cycle, he gains from the choices of both Archie and Billie before making his own decision.
- If Archie and Billie are unable to make a move (meaning they cannot choose a larger $n$ than previously chosen and have $2n$ pebbles), they lose.

- Charlie always has an advantage as he gets to benefit from two preceding loss functions (first by Archie's then by Billie's subtraction), which accumulate to hasten their depletion before his resources end.

Thus, Charlie has a strategy to avoid losing, assuming they optimally choose when $n$ runs closer to depleting Archie and Billie's resources:

$$\boxed{\text{Charlie}}$$