Solution 1. The possible values are 1, 2, and 2024.

Construction. The function $f(x) = 1$ for all $x \in \mathbb{N}$ works. Also, $f(x) = 1$ for all $x \neq 2024$ and $f(2024) = 2$, works. Finally, $f(x) = x$ for all $x \in \mathbb{N}$ works as well.
It remains to show these are the only possible values for $f(2024)$.

Proof. Denote $\text{Im}(f) = \{f(x) \mid x \in \mathbb{N}\}$. The cardinal set $\{1\}$ gives $f(1) = 1$. Consider the following two cases:

* $\text{Im}(f)$ is unbounded. Fix any $n \in \mathbb{N}$, with $n > 1$. Pick $n-1$ distinct integers $k_1, \dots, k_{n-1}$ such that $f(k_i) \notin \{n, f(n)\}$ and $f(k_i)$ are all pairwise distinct, for $1 \le i < n$. Then $\{n, k_1, \dots, k_{n-1}\}$ is a cardinal set. Then $\{f(n), f(k_1), \dots, f(k_{n-1})\}$ is a cardinal set with $n$ distinct elements, so $n$ lies in this set, hence $f(n) = n$. This gives the identity function.

* $\text{Im}(f)$ is bounded. Suppose $f(x) \le M$ for all $x \in \mathbb{N}$ and some integer $M > 0$.

Claim. For any integer $a$ satisfying $1 \le a \le M$, if there are infinitely many integers $n \in \mathbb{N}$ such that $f(n) = a$, then $a = 1$.

Proof. Let $b > 1$ be one of the integers with $f(b) = a$. Consider $b-1$ other integers $c_1, \dots, c_{b-1}$, such that $f(c_i) = a$ for $1 \le i < b$, and $c_i$ are all pairwise distinct. Then $\{b, c_1, \dots, c_{b-1}\}$ is a cardinal set, so the image set, which consists of the singleton $\{a\}$ is cardinal, hence $a = 1$. $\square$

So for every $2 \le m \le M$, there are only finitely many integers $x$ such that $f(x) = m$. Thus, there exists an integer $N > 1$ such that for all $n \ge N$, $f(n) = 1$. Now for every $1 < l < N$, consider the cardinal set $\{l, N+1, N+2, \dots, N+l-1\}$. Then the image set consists of $\{1, f(l)\}$, which can be cardinal only when $f(l) = 1$ or $f(l) = 2$.

By the above reasoning, $f(2024)$ can only be 1, 2, or 2024, each of which occurs as an example. $\square$

Solution 2. We present a second proof of the fact that the proposed values are the only possibilities. Considering the singleton cardinal set $\{1\}$, we see that $f(1) = 1$. The cardinal set $\{1, 2\}$ gets mapped to $\{1, f(2)\}$, so $f(2)$ must be 2 or 1.

Case 1. Suppose $f(2) = 1$. Now $\{2, 2024\}$ is a cardinal set, and therefore so is $\{1, f(2024)\}$. This means $f(2024)$ is 1 or 2.

Case 2. Suppose $f(2) = 2$. The cardinal set $f(\{1, 2, 3\}) = \{1, 2, f(3)\}$ shows that $f(3) \in \{1, 2, 3\}$, but the cardinal set $f(\{2, 3\}) = \{2, f(3)\}$ proves $f(3)$ cannot be 2. Thus there are two sub-cases:

2.1. $f(3) = 1$. Then the set $\{1, 3, 2024\}$ is cardinal, hence so is $\{1, f(2024)\}$, implying, as before, $f(2024) \in \{1, 2\}$.

2.2. $f(3) = 3$. In this case, we show via induction that $f(n) = n$ for all $n \in \mathbb{N}$.

The base cases $n = 1, 2, 3$ are already known. Now consider $n \ge 4$, and assume $f(k) = k$ for all $k < n$. Consider the cardinal $f(\{1, 2, \dots, n\}) = \{1, 2, \dots, n-1, f(n)\}$ which implies $f(n) \in \{1, 2, \dots, n\}$.

However, consider the $n-1$-element cardinal set $\{1, 2, \dots, n\} \setminus \{n-2\}$. For its image to be cardinal $f(n)$ cannot equal any number in $\{1, 2, \dots, n-1\} \setminus \{n-2\}$; else its cardinality would be $n-2$, which isn't in the set. So $f(n) \in \{n-2, n\}$.

Finally, consider the $n-2$-element set $\{1, 2, \dots, n\} \setminus \{n-1, n-3\}$. If $f(n) = n-2$, its image would only have $n-3$ elements, and thus would not be cardinal. So we conclude that $f(n) = n$ and the induction is complete. In particular, $f(2024) = 2024$.