Balkan MO Shortlist

Let $S$ be a finite set with $|S| = n$. We are asked to find the maximum size of the image of a function $f$ in the set $\mathcal{F}$, where $\mathcal{F}$ is the set of all functions $f : \mathcal{P}(S) \to \mathbb{R}$ satisfying the condition that for all subsets $X, Y \subseteq S$, we have:
$$f(X \cap Y) = \min(f(X), f(Y)).$$
The objective is to maximize $|\text{Im}(f)|$, the number of distinct values taken by $f$.

### Step-by-step Analysis

1. Understanding the Condition:
The condition $f(X \cap Y) = \min(f(X), f(Y))$ implies that for any subsets $X$ and $Y$ of $S$, if $X \cap Y = \emptyset$, we should have $f(\emptyset) = \min(f(X), f(Y))$. Therefore, $f(\emptyset)$ is the smallest value in the image of $f$.

2. Construction of the Function:
Consider constructing $f$ based on the size of subsets. Define $f(T)$ for every subset $T \subseteq S$ based on its size:
$$f(T) = |T|.$$

3. Image of the Function:
Under this construction:
- $f(\emptyset) = 0$.
- For any subset $T \subseteq S$ with $|T| = k$, $f(T) = k$.

As $T$ can be any subset, the value of $f(T)$ can range from $0$ to $n$, where $n = |S|$.

4. Size of the Image Set:
The image of $f$, $\text{Im}(f)$, contains all integers from $0$ to $n$, inclusive. Therefore, $|\text{Im}(f)| = n + 1$.

5. Verification:
Verify that this satisfies the condition $f(X \cap Y) = \min(f(X), f(Y))$. For subsets $X$ and $Y$:
$$f(X \cap Y) = |X \cap Y| = \min(|X|, |Y|) = \min(f(X), f(Y)).$$

In conclusion, the maximum size $|\text{Im}(f)|$ for functions $f \in \mathcal{F}$ is indeed:
$$\boxed{n + 1}.$$
This represents the distinct non-negative integer values from $0$ up to the size of the set $S$, fulfilling the function condition defined in $\mathcal{F}$.