A finite set $\mathcal{S}$ of positive integers is called cardinal if $\mathcal{S}$ contains the integer $|\mathcal{S}|$ where $|\mathcal{S}|$  denotes the number of distinct elements in $\mathcal{S}$. Let $f$ be a function from the set of positive integers to itself such that for any cardinal set $\mathcal{S}$, the set $f(\mathcal{S})$ is also cardinal. Here $f(\mathcal{S})$ denotes the set of all integers that can be expressed as $f(a)$ where $a \in \mathcal{S}$. Find all possible values of $f(2024)$
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Proposed by Sutanay Bhattacharya
1. **Understanding the Problem:**
   We need to find all possible values of \( f(2024) \) given that \( f \) is a function from the set of positive integers to itself, and for any cardinal set \( \mathcal{S} \), the set \( f(\mathcal{S}) \) is also cardinal. A set \( \mathcal{S} \) is cardinal if it contains the integer \( |\mathcal{S}| \), where \( |\mathcal{S}| \) denotes the number of distinct elements in \( \mathcal{S} \).

2. **Initial Observations:**
   - For \( \mathcal{S} = \{1\} \), \( f(\{1\}) = \{f(1)\} \) must be cardinal, implying \( f(1) = 1 \).
   - For \( \mathcal{S} = \{1, 2\} \), \( f(\{1, 2\}) = \{1, f(2)\} \) must be cardinal, implying \( f(2) = 1 \) or \( f(2) = 2 \).

3. **Defining a Set \( T \):**
   Define \( T = \{k : f(k) = 2\} \). We need to show that \( |T| \leq \min(T) - 1 \).

4. **Proof of Lemma 1:**
   Suppose \( |T| \geq \min(T) \). Choose \( V \subset T \) such that \( \min(T) \in V \) and \( |V| = \min(T) \). Since \( V \) is cardinal, \( f(V) = \{2\} \) must also be cardinal, which is a contradiction because \( \{2\} \) is not cardinal. Hence, \( |T| \leq \min(T) - 1 \).

5. **Case 1: \( f(2) = 1 \):**
   - The codomain of \( f \) is restricted to \( \{1, 2\} \).
   - Suppose there exists \( x \) such that \( f(x) \geq 3 \). Then \( f(\{2, x\}) = \{1, f(x)\} \) must be cardinal, which is a contradiction because \( \{1, f(x)\} \) is not cardinal. Hence, \( f(x) \) cannot be greater than 2.

6. **Case 2: \( f(2) = 2 \):**
   - By Lemma 1, \( f(3) \neq 2 \).
   - For \( \mathcal{S} = \{1, 2, 3\} \), \( f(\{1, 2, 3\}) = \{1, 2, f(3)\} \) must be cardinal, implying \( f(3) = 1 \) or \( f(3) = 3 \).

7. **Subcase 1: \( f(3) = 1 \):**
   - The codomain of \( f \) is restricted to \( \{1, 2\} \).
   - Suppose there exists \( x \) such that \( f(x) \geq 3 \). Then \( f(\{1, 3, x\}) = \{1, f(x)\} \) must be cardinal, which is a contradiction because \( \{1, f(x)\} \) is not cardinal. Hence, \( f(x) \) cannot be greater than 2.

8. **Subcase 2: \( f(3) = 3 \):**
   - We claim that \( f(n) = n \) for all \( n \in \mathbb{N} \).
   - Base cases: \( f(1) = 1 \), \( f(2) = 2 \), and \( f(3) = 3 \).
   - Inductive step: Assume \( f(i) = i \) for all \( i \in \{1, 2, \dots, k\} \).
   - **Lemma 2:** \( f(k+1) \leq k+1 \).
     - Suppose \( f(k+1) > k+1 \). Then \( f(\{1, 2, \dots, k, k+1\}) = \{1, 2, \dots, k, f(k+1)\} \) must be cardinal, which is a contradiction because \( \{1, 2, \dots, k, f(k+1)\} \) is not cardinal.
   - If \( f(k+1) = l \) where \( l \in \{1, 2, \dots, k\} \):
     - If \( l = 1 \), \( f(\{1, 2, \dots, k-1, k, k+1\}) = \{1, 2, \dots, k-1, k\} \) is not cardinal.
     - If \( l = 2 \), \( f(\{2, k+1\}) = \{2\} \) is not cardinal.
     - If \( 3 \leq l \leq k \), \( f(\{1, 2, \dots, l-2, l, k+1\}) = \{1, 2, \dots, l-2, l\} \) is not cardinal.
   - Therefore, \( f(k+1) = k+1 \) is forced, and by induction, \( f(n) = n \) for all \( n \in \mathbb{N} \).

9. **Conclusion:**
   - If \( T = \emptyset \), then \( f(n) = 1 \) for all \( n \in \mathbb{N} \), and \( f(2024) = 1 \).
   - If \( T = \{2024\} \), then \( f(n) = 1 \) for all \( n \in \mathbb{N} - \{2024\} \), and \( f(2024) = 2 \).
   - If \( f(n) = n \) for all \( n \in \mathbb{N} \), then \( f(2024) = 2024 \).

The final answer is \( \boxed{ f(2024) \in \{1, 2, 2024\} } \).