, 2024

Consider a sweet $n$-configuration $\langle A_{i,j} \rangle_{1\le i,j\le n}$ with $A_{n,n} \subset \{1, 2, \dots, k\}$. For any $x \in \{1, 2, \dots, k\}$ and $i \in \{1, 2, \dots, n\}$ define
$$p_x(i) = |\{j: x \in A_{i,j}\}|.$$
Since $A_{i,j} \subseteq A_{i,j+1}$ for all suitable $i, j$, the set $\{j: x \in A_{i,j}\}$ consists of $p_x(i)$ largest elements of $\{1, 2, \dots, n\}$. Since $A_{i,j} \subseteq A_{i+1,j}$ for all suitable $i, j$, the function $p_x: \{1, 2, \dots, n\} \to \{0, 1, 2, \dots, n\}$ is nondecreasing. Therefore every sweet $n$-configuration determines a family $\langle p_x \rangle_{1\le x\le k}$ of nondecreasing functions $p_x: \{1, 2, \dots, n\} \to \{0, 1, \dots, n\}$. Conversely, every such a family determines a sweet $n$-configuration $\langle A_{i,j} \rangle_{1\le i,j\le n}$ with $A_{n,n} \subset \{1, 2, \dots, k\}$ in the following way: $A_{i,j} = \{x \in \{1, 2, \dots, k\}: j \ge n+1-p_x(i)\}$. Therefore $f(n,k) = g(n)^k$ where $g(n)$ is the number of nondecreasing functions $p: \{1, 2, \dots, n\} \to \{0, 1, \dots, n\}$.

Using the stars-and-bars method, there is a bijection between the family of nondecreasing functions $p: \{1, 2, \dots, n\} \to \{0, 1, \dots, n\}$ and the set of sequences consisting of $n$ stars and $n$ bars. The bijection is given by
$$p \to \underbrace{**\dots*}_{p(1)} | \underbrace{**\dots*}_{p(2)-p(1)} | \underbrace{**\dots*}_{p(3)-p(2)} | \dots | \underbrace{**\dots*}_{p(n)-p(n-1)} | \underbrace{**\dots*}_{n-p(n)}$$
Thus $g(n) = \binom{2n}{n}$.

The problem boils down to determining which of the numbers
$$\binom{2n}{n}^{n^2}, \quad \binom{2n^2}{n^2}^n,$$
where $n = 2024$, is larger. Note that
$$\begin{aligned} \binom{2n^2}{n^2} &= \frac{\prod_{i=1}^{n^2} (n^2 + i)}{\prod_{i=1}^{n^2} i} = \prod_{i=1}^{n^2} \left(\frac{n^2 + i}{i}\right) = \prod_{j=0}^{n-1} \prod_{i=1}^{n} \frac{n^2 + jn + i}{jn + i} > \prod_{j=0}^{n-1} \left(\frac{n^2 + jn + n}{jn + n}\right)^n = \\ &= \prod_{j=0}^{n-1} \left(\frac{n + j + 1}{j + 1}\right)^n = \left(\prod_{j=1}^{n} \frac{n+j}{j}\right)^n = \left(\binom{2n}{n}\right)^n \end{aligned}$$
and therefore
$$\binom{2n^2}{n^2}^n > \binom{2n}{n}^{n^2}.$$

Remark: A sketch of a slightly different way of thinking about $f(n,k) = \binom{2n}{n}^k$: Consider an $n \times n$ table. In a cell with coordinates $(i, j)$, list all the elements of the set $A_{i,j}$. Fix an element $x \in \{1, \dots, k\}$ and consider the cells that contain the number $x$. By the condition, those cells form a region closed under making a step right and making a step up. Such regions are delimited by grid paths that start at $[0, n]$, end at $[n, 0]$, and only steps right or down. There are $\binom{2n}{n}$ possible paths for each $x$, thus $f(n,k) = \binom{2n}{n}^k$.