IMO 2J, Independent Study 1

Clearly $f(x) = x$ is one solution. Furthermore, for any set $A = \{a_1, a_2, \dots, a_{2024}\}$ consisting of 2024 distinct positive integers, any function $f$ satisfying
$$\begin{cases} f(a_i) = a_{i-1}, & \text{where } a_0 = a_{2024}, \\ f(x) \in A, & \text{for all } x \notin A, \end{cases}$$
also satisfies the condition of the problem. Let us prove that the above are all the solutions.

Using all positive integers as points, and connecting $x$ to $f(x)$, we form a directed graph $G$ (self-loops allowed). Note that if $x$ can reach 2025 points, then any $A$ containing $x$ and $f(x)$ would have $|S_A| \ge 2025$, a contradiction. This means that starting from $x$, we must enter a cycle within 2024 steps, and the size of this cycle is at most 2024.

Now, consider all points in $G$ that lie on cycles. Let us consider the size of $C$.

Case 1. $|C| > 2024$:
Note that in this case there cannot be any $x \in C$ such that $f(x) \ne x$; otherwise, take any $A \subset C$ containing $x$ but not $f(x)$, then $A \subset S_A$ (note that all elements of $A$ lie on cycles) and $f(x) \in S_A$, so that $A \cup \{f(x)\} \subset S_A \Rightarrow |S_A| \ge |A| + 1 > 2024$, a contradiction.

Hence we have $f(x) = x$ for all $x \in C$. Now, if there is any $y \notin C$, since $y$ must reach $C$ within 2024 steps, there exists $z \notin C$ such that $f(z) \in C$. If we take $A = \{z, f(z), b_1, \dots, b_{2022}\}$, where $b_i$ are 2022 elements of $C$ different from $f(z)$, then $S_A = A - \{z\}$, so that $|S_A| = |A| - 1 < 2024$, a contradiction. Hence $C = \mathbb{N}$, that is, $f(x) = x$.

Case 2. $|C| \le 2024$:
Since all $x \in \mathbb{N}$ must reach $C$ within 2024 steps, and $C$ is a finite set, there exists $y \in \mathbb{N}$ such that $f^{-1}(y) := \{x : f(x) = y\}$ is an infinite set. Now taking $A \subset f^{-1}(y)$, we know that $Y := \{y, f(y), \dots, f^{(2023)}(y)\}$ has $|Y| = 2024$; in other words, $y$ lies on a cycle $Y$ of size exactly 2024.

Now, for any $x \in \mathbb{N}$, take $A$ satisfying $x \in A$ and $A \cap f^{-1}(y) \neq \emptyset$. Note that in this case $Y \subset S_A$, forcing $Y = S_A$ (otherwise $|S_A| > |Y| = 2024$), and hence also $f(x) \in Y$. This means that all points not on $Y$ must reach $Y$ in one step, which is the second possible $f$.