Taiwan IMO Selection Camp, 2023

For $Y \subset X$ and any function $f$ with domain $X$, let us define $f(Y) = \sum_{x \in Y} f(x)$; then $f$ is nice is equivalent to saying that $f(X_i)$ attains its maximum at a unique index $i \in \{1, 2, \dots, m\}$.
Let $\mathcal{G}$ be the set of all functions $g : X \to \{1, 2, \dots, n\}$; note that $|\mathcal{G}| = n^n$. Now, for each $g \in \mathcal{G}$, take any index $\ell$ maximizing $g(X_\ell)$, and define
$$f_g(x) := \begin{cases} g(x) + 1, & x \in X_\ell, \\ g(x), & x \notin X_\ell. \end{cases}$$
If we can show that (1) $f_g$ is nice and (2) the $f_g$ are all distinct, then this proves that there are at least $n^n$ nice functions.

(1) **$f_g$ is nice** Note that $f_g(X_i) = g(X_i) + |X_i \cap X_\ell|$ holds for all $i$. Let us show that $f_g(X_i)$ attains its maximum at the unique index $\ell$: take any $i \neq \ell$. Note that we must have $X_\ell \not\subset X_i$ (otherwise $g(X_\ell) < g(X_i)$, contradicting the maximality of $g(X_\ell)$), and this implies $|X_\ell| > |X_i \cap X_\ell|$, so
$$f_g(X_\ell) = g(X_\ell) + |X_\ell| \ge g(X_i) + |X_\ell| > g(X_i) + |X_i \cap X_\ell| = f_g(X_i).$$
Thus $f_g(X_i)$ attains its maximum at the unique index $\ell$, that is, $f_g$ is nice.

(2) **The $f_g$ are all distinct** Note that the previous part shows that $f_g(X_\ell)$ is the unique maximum, so we have
$$g(x) := \begin{cases} f_g(x) - 1, & x \in X_\ell, \\ f_g(x), & x \notin X_\ell. \end{cases}$$
This shows that $f_g$ and $g$ are in one-to-one correspondence, hence the $f_g$ are all distinct.