can i have the pdf file

In fact, when $n$ is even, the best $C$ is $n$ The solution of $n$ even is similar to $n$ odd

I'm the author of this problem (and part of the solution). Hope you enjoyed it. I would love to see different solutions!

Call a nonempty subset of unlocked chests a state. For each state $S \subseteq \{1, \dots, 2023\}, S \neq \emptyset$ and $x \in S$, let $f(S, x)$ be the number of times that the fairy has chosen to lock chest $x$ while in state $S$, and let $g(S, x)$ be the number of times that Elisa has chosen to add a gem to chest $x$ while in state $S$. Claim: Elisa has a strategy ensuring at all times that for all $S, x$, $|f(S, x) - g(S, x)| \le 1$. Proof: The strategy is as follows: when Elisa is in some state $S$, if this is the first time $S$ has been reached, then Elisa chooses an arbitrary unlocked chest to add a gem; otherwise, Elisa adds a gem to the chest that the fairy locked on the previous occurrence of state $S$. Then, for each state $S$ and chest $x \in S$, there is a bijection between gems added to $x$ in state $S$ and occasions where $x$ was locked while in state $S$, with the exceptions of at most one gem added on the first occurrence of $S$ and the most recent time that $x$ was locked. $\square$ Now we claim that $C = 2^{2023} + 1$ works. For each chest $x$, the number of gems in $x$ is given by $\sum_{S \subseteq \{1, \dots, 2023\}, x \in S} f(S, x)$. Then for two chests $x, y$, the difference in the number of gems is \begin{align*} &\sum_{S \subseteq \{1, \dots, 2023\}, x \in S} f(S, x) - \sum_{S \subseteq \{1, \dots, 2023\}, y \in S} f(S, y) \\ &\le \sum_{S \subseteq \{1, \dots, 2023\}, x \in S} g(S, x) + |f(S, x) - g(S, x)| - \sum_{S \subseteq \{1, \dots, 2023\}, y \in S} g(S, y) - |g(S, y) - f(S, y)| \\ &\le 2^{2023} + \sum_{S \subseteq \{1, \dots, 2023\}, x \in S} g(S, x) - \sum_{S \subseteq \{1, \dots, 2023\}, y \in S} g(S, y). \end{align*} This bound is $2^{2023}$ plus the difference in the number of times chest $x$ and chest $y$ were locked, which cannot exceed $1$.

Claim: It is always possible for Elisa to place the gems such that if we let $N_i$ be the number of gems in the chest with the $i$-th most gems, that after $m$ rounds where a round is $n$ moves by Elisa the sequence \[m+\lfloor\frac{n}{2}\rfloor, m+\lfloor\frac{n}{2}\rfloor-1, \dots, m+\lfloor\frac{n}{2}\rfloor-n \succ N_1, N_2, \dots, N_n\] Proof: To do this Elisa can always play in the chest with the smallest number of gems that is avaliable. I will show this inductively, suppose that after $k$ moves in the $m$-th round \[m+\lfloor\frac{n}{2}\rfloor, m+\lfloor\frac{n}{2}\rfloor-1, \dots, m+\lfloor\frac{n}{2}\rfloor-k+1, m+\lfloor\frac{n}{2}\rfloor-k+1, \dots, m+\lfloor\frac{n}{2}\rfloor-n+1 \succ N_1, N_2, \dots, N_n\]I will show that by choosing the smallest avaliable option that \[m+\lfloor\frac{n}{2}\rfloor, m+\lfloor\frac{n}{2}\rfloor-1, \dots, m+\lfloor\frac{n}{2}\rfloor-k+2, m+\lfloor\frac{n}{2}\rfloor-k+2, \dots, m+\lfloor\frac{n}{2}\rfloor-n+1 \succ N_1, N_2, \dots, N_n\], on the $k+1$-th move, as we choose the smallest and there are only $k$ closed chests, we choose one of the $k+1$ smallest of the options, thus the result holds and thus after each round the claim holds.

OG! We claim $C=2022$ suffices. Strategy for Elsa: Elsa always adds a gem to the unlocked chest with THE smallest number of gems. Proof that this strategy works: Will write later Remark: In general if 2023 is replaced by $k$, $C= 2(\lfloor \frac{k}{2} \rfloor)$ works