A family $\mathcal{F}$ of sets is called [i]really neat[/i] if for any $A,B\in \mathcal{F}$, there is a set $C\in \mathcal{F}$ such that $A\cup B = A\cup C=B\cup C$. Let
$$f(n)=\min \left\{ \max_{A\in \mathcal{F}} |A| \colon \mathcal{F} \text{ is really neat and } |\cup \mathcal{F}| =n\right\} .$$
Prove that the sequence $f(n)/n$ converges and find its limit.
1. **Prove that \( f(n) \ge \frac{n}{2} \):**

   Suppose \(\mathcal{F}\) is a family of sets such that \(\left|\bigcup \mathcal{F}\right| = n\). Note that \(A \cup B = A \cup C = B \cup C\) implies \(A \bigtriangleup B \subseteq C \subseteq A \cup B\), where \(\bigtriangleup\) denotes the symmetric difference of two sets. For any two elements \(A, B \in \mathcal{F}\), there exists \(C \in \mathcal{F}\) with \(C \supseteq A \bigtriangleup B\).

   Denote \(F = \bigcup \mathcal{F}\). Let \(A_0 \in \mathcal{F}\) be a set with the maximal number of elements. Consider the family \(\mathcal{F}_1 := \{A \cap (F \setminus A_0) : A \in \mathcal{F}\}\). Let \(A_1 \in \mathcal{F}_1\) be the set with the maximal number of elements.

   **Claim:** \( |A_1| \le \frac{|A_0|}{2} \).

   **Proof of Claim:**
   Suppose, for the sake of contradiction, that \( |A_1| > \frac{|A_0|}{2} \). Let \(A'_1 \in \mathcal{F}\) be the corresponding set for which \(A_1 = A'_1 \cap (F \setminus A_0)\). Then \( |A'_1 \cap A_0| \ge |A_1| \) since otherwise there is a set \(B \in \mathcal{F}\) with \(B \supseteq A \bigtriangleup A'_1\) and \( |B| > |A_0| \), contradicting the maximality of \(A_0\).

   Hence, \( |A'_1| \ge 2|A_1| > |A_0| \), which again contradicts the maximality of \(A_0\). \(\blacksquare\)

   Note that \(\mathcal{F}_1\) satisfies the same condition as \(\mathcal{F}\). Thus, we can proceed with the same argument as above with \(\mathcal{F}_1, A_1\) instead of \(\mathcal{F}, A_0\) and find \(\mathcal{F}_2, A_2\) with \( |A_2| \le \frac{|A_1|}{2} \). In this way, we construct a family of disjoint sets \(A_j\) with \( |A_{j+1}| \le \frac{|A_j|}{2} \) for \( j = 0, 1, \dots \), and moreover, they must cover all \( n \) elements of \(\bigcup \mathcal{F}\). This means \( |A_0| > \frac{n}{2} \), hence \( f(n) > \frac{n}{2} \).

2. **Provide an example of such a family:**

   Suppose \( n = 2^k \) and \( F = \{0, 1, \dots, n-1\} \). Let \( A_i, i = 1, 2, \dots, k \) be the set of those elements of \( F \) whose binary representation has the digit \( 1 \) in the \( i \)-th place. Obviously, \( |A_i| = \frac{n}{2} \) for \( i = 1, 2, \dots, k \).

   Next, starting from \(\{A_i : i = 1, 2, \dots, k\}\), we generate the smallest family \(\mathcal{F}\) which is closed under taking the symmetric difference. The size of any element of \(\mathcal{F}\) is \(\frac{n}{2}\) and they cover all the set \( F \) except \( 0 \). Thus, \( |\cup \mathcal{F}| = n - 1 \). Adding the element \( 0 \) to all sets in \(\mathcal{F}\), we obtain a family covering \( F \), each set of size \(\frac{n}{2} + 1\).

   In the general case, when \( n = 2^{k_1} + 2^{k_2} + \dots + 2^{k_s} \) with \( k_1 > k_2 > \dots > k_s \ge 0 \), we apply the above construction to some disjoint sets \( F_i, i = 1, 2, \dots, s \), each consisting of \( 2^{k_i} \) elements, thus obtaining families \(\mathcal{F}_1, \mathcal{F}_2, \dots, \mathcal{F}_s\). Then we consider
   \[
   \mathcal{F} := \{A : A = \bigcup_{i=1}^s A_i, A_i \in \mathcal{F}_i, i = 1, 2, \dots, s\}.
   \]
   It is easily checked that \(\mathcal{F}\) satisfies the requirements and the size of each set in it is at most \(\frac{n}{2} + \left\lfloor \log_2 n \right\rfloor + 1\).

   Thus, it has been proven that
   \[
   \frac{n}{2} < f(n) \le \frac{n}{2} + \left\lfloor \log_2 n \right\rfloor + 1,
   \]
   and the result follows.

The final answer is \( \boxed{ \frac{1}{2} } \).