For a subset $Y \subseteq X$, we write $f(Y)$ for $\sum_{y \in Y} f(y)$. Note that a function $f: X \rightarrow \{1, \ldots, n+1\}$ is nice, if and only if $f\left(X_{i}\right)$ is maximized by a unique index $i \in \{1, \ldots, m\}$. We will first investigate the set $\mathcal{F}$ of functions $f: X \rightarrow \{1, \ldots, n\}$; note that $|\mathcal{F}|=n^{n}$. For every function $f \in \mathcal{F}$, define a corresponding function $f^{+}: X \rightarrow \{1,2, \ldots, n+1\}$ in the following way: Pick some set $X_{l}$ that maximizes the value $f\left(X_{l}\right)$.

- For all $x \in X_{l}$, define $f^{+}(x)=f(x)+1$.
- For all $x \in X \backslash X_{l}$, define $f^{+}(x)=f(x)$.

Claim. The resulting function $f^{+}$ is nice.

Proof. Note that $f^{+}\left(X_{i}\right)=f\left(X_{i}\right)+\left|X_{i} \cap X_{l}\right|$ holds for all $X_{i}$. We show that $f^{+}\left(X_{i}\right)$ is maximized at the unique index $i=l$. Hence consider some arbitrary index $j \neq l$. Then $X_{l} \subset X_{j}$ is impossible, as this would imply $f\left(X_{j}\right)>f\left(X_{l}\right)$ and thereby contradict the choice of set $X_{l}$; this in particular yields $\left|X_{l}\right|>\left|X_{j} \cap X_{l}\right|$.

$$f^{+}\left(X_{l}\right)=f\left(X_{l}\right)+\left|X_{l}\right| \geqslant f\left(X_{j}\right)+\left|X_{l}\right|>f\left(X_{j}\right)+\left|X_{j} \cap X_{l}\right|=f^{+}\left(X_{j}\right)$$

The first inequality follows since $X_{l}$ was chosen to maximize the value $f\left(X_{l}\right)$. The second (strict) inequality follows from $\left|X_{l}\right|>\left|X_{j} \cap X_{l}\right|$ as observed above. This completes the proof of the claim.

Next observe that function $f$ can be uniquely reconstructed from $f^{+}$: the claim yields that $f^{+}$ has a unique maximizer $X_{l}$, and by decreasing the value of $f^{+}$ on $X_{l}$ by 1, we get we can fully determine the values of $f$. As each of the $n^{n}$ functions $f \in \mathcal{F}$ yields a (unique) corresponding nice function $f^{+}: X \rightarrow \{1,2, \ldots, n+1\}$, the proof is complete.