A refreshing combi. Somewhat unintuitively (or maybe not), only $C$ can guarantee not to lose. Firstly, we show $C$ guarantees not to lose with the following strategy: Let $A$ make the first move and then mimic $B$ by always playing one more pebble than him. This works as after the first move, $C$ has $2025$ pebbles, and we'll prove that after every move, $A$ makes this remains true. Additionally, we'll show that $C$ can make the desired move. Assume $B$ plays $n$. Then $2n\leq 3\cdot 2024-2025$, so $n$ is at most $2023$. Then $C$ has at least $2025+n\geq 2n+2$ pebbles, so he can make his move of $n+1$. Also, $A$ must add at least $n+2$ to $C$'s pile so Charlie doesn't lose pebbles overall in these three moves because $n+(n+2)=2(n+1)$. Thus, $C$ can't lose. We now show that $C$ can play in such a way that he chooses which $A$ and $B$ loses. We presume that both $A$ and $B$ play so that they guarantee that the other two players can't force them to lose at each move of theirs. With this in mind, we'll show that $A$ plays $n=1$, then $B$ plays $n=2$, and so on. Denote by $p_k$ the position where $A$ has $2024+3k$ pebbles, $B$ has $2024$ pebbles and $C$ has the remaining $2024-3k$ pebbles. In particular, the starting position is trivially $p_0$. We can show that in a $p_k$, whatever $A$ plays, $B$ and $C$ can make moves in such a way that after these three moves, we emerge in $p_{k+1}$ if $k\leq 673$. Indeed, if $A$ chooses $n$, $B$ and $C$ can commit to choosing $n+1$ and $n+2$. Note that $2024-3k\geq 5$, so $2024+3k\leq 4043$. This implies $n\leq 2021$. Therefore, $2(n+1)<2024+n$, and $B$ can play $n+1$. After that $C$ has $n+(n+1)+(2024-3k)\geq 2n+6>2(n+2)$ pebbles, so he can play $n+2$. Therefore, if $k\leq 673$, $B$ and $C$ can play so that $p_k$ becomes $p_{k+1}$. However, $A$ would like not to lose. This is only possible in the position $p_{674}$ and only if $A$ can play $2023$ in that position. Therefore, going back, $A$ can't play $n>3k+1$ at a position $p_k$ because then $B$ and $C$ can force him to have to play $n>2023$ in $p_{674}$ or lose earlier, contradiction. This implies that $A$ must play $n=1$ in $p_0$. Now, on to the first move of $B$, assume it's not $n=2$. Then, with a similar strategy (here the roles are switched cyclically; $C$ plays $B$, $A$ plays as $C$, and $A$ plays as $C$), we'll show $B$ loses if $C$ and $A$ team up. Also, $B$ has $2025$ pebbles initially, so the inequalities required are met. We go into positions $(2025+3k, 2025, 2022-3k)$ for $(B, C, A)$, respectively, and if $B$ dares to increase $n$ by more than three between every three moves, then the $n\to n+1\to n+2$ strategy fails him. Therefore, $B$ must play $3$, then $6$ and so on. In the end, we have: \[(4044, 2025, 3) \to (0, 4047, 2025) \to (2023, 1, 4048) \to (4047, 2025, 0)\text{ and $B$ loses.}\]In summary, to avoid losing, $A$ must therefore play $n=1$, and then $B$ must play $n=2$. Then $C$ may play $n=3$, and the steps continue consecutively using the same logic. We now consider what happens and how $C$ can choose who loses. We have: \[(4043, 2024, 5) \to (3, 4044, 2025) \to (2024, 2, 4046)\]Now if $C$ plays $n=2023$, $A$ loses, and if $C$ plays $n=2022$, $B$ loses. This completes the proof.

New JBMO record: this problem has been solved completely by only one contestant (Nina Susic from Serbia). Let us hope that in a future edition there will not be a problem with zero solutions.

Alternative Solution for $C$: Lemma $1$: The game ends in at most $2023$ turns. Proof $1$: Let's prove that $B_i\leq B_{i-1}-3$ where $B_j$ is the number of pebbles $B$ has after his $j.$ move. \[(a_i,b_i,c_{i-1})\rightarrow (a_i+x,b_i+x,c_{i-1}-2x)\rightarrow (a_i+x-2y,b_i+x+y,c_{i-1}-2x+y)\rightarrow (a_i+x-2y+z,b_i+x+y-2z,c_{i-1}-2x+y+z)\]Hence $b_{i+1}=b_i+(x+y-2z)$ where $z>y>x$. $b_{i+1}=b_i+(x-z)+(y-z)\leq b_i+(-2)+(-1)=b_i-3$ Suppose that the game did not end in $2023$ turns. Then, $B$ makes at least $675$ moves. \[B_{675}\leq B_0-3.675=2024-2025=-1\]Which is impossible.$\square$ Lemma $2$: If $C$ loses after some number of turns, then $2.(\text{number of pebbles of C)}-(\text{number of pebbles of B})\leq 2$ must hold after a move of $A$. Proof $2$: Assume that $2c-b\geq 3$ after each move of $A$. \[(a,b,c)\rightarrow (a+k,b-2k,c+k)\]If $C$ cannot make a move, then $c+k<2(k+1)\iff c+k\leq 2k+1\iff c\leq k+1$. Also $b-2k\geq 0\iff b\geq 2k$. Thus, $\frac{b}{2}\geq k\geq c-1\implies 1\geq c-\frac{b}{2}\iff2\geq 2c-b$ which gives a contradiction.$\square$ Now let's prove that if $c_i=b_i+1,$ then $C$ doesn't lose the game. We will show that $2c-b\geq 3$ after $A'$s moves during $\leq 2023$ turns. Let $f(i)=2c-b$ after $i.$ move. \[(a,b,c)\rightarrow (a+x,b-2x,c+x)\rightarrow (a+2x+1,b-x+1,c-x-2)\rightarrow (a+2x+1-2y,b-x+1+y,c-x-2+y)\]$f(3i+1)=2c-2x-4+2y-b+x-1-y=(2c-b)+(y-x)-5\geq (2c-b)-3=f(3i-2)+3$ Hence $f(3k+1)\geq f(3k-2)-3$. Since $f(1)\geq 2.2025-2025=2025,$ we have $f(3k+1)\geq f(1)-3k\geq 2025-2022=3$ which shows that $f\geq 3$ during the game as desired.$\blacksquare$

Let A,B and C be the number of pebbles in pile of the players Archie Billie and Charlie respectively we since there are $1012$ even integers less than $2024$ we know that if all the players play perfectly game will end in $1012$ steps since $1012$ is $1(\bmod 3)$ we know Alice made the last move.If so,Billie loses but if do the calculations as follows; $A=A_0-2(\sum n_{3k+1})+\sum n_{3k+2}+\sum n_{3k}$ $B=B_0-2(\sum n_{3k+2})+\sum n_{3k+1} +\sum n_{3k}$ $C=C_0-2(\sum n_{3k})+\sum n_{3k+1} +\sum n_{3k+2}$ Since all the numbers $A,B$ and $C$ should be positive if we sum $A+B+C$ is also a positive integer.But if you sum all the $RHS$’s you will get $-1012.1013$ which is a negative integer.From this we can say that $A$ cant make a move and the player who makes a move before $A$ wins and that is Charlie