In a competition there are $18$ teams and in each round $18$ teams are divided into $9$ pairs where the $9$ matches are played coincidentally. There are $17$ rounds, so that each pair of teams play each other exactly once. After $n$ rounds, there always exists $4$ teams such that there was exactly one match played between these teams in those $n$ rounds. Find the maximum value of $n$.
1. **Understanding the Problem:**
   We have 18 teams, and in each round, they are divided into 9 pairs to play 9 matches. There are 17 rounds in total, and each pair of teams plays exactly once. We need to find the maximum value of \( n \) such that after \( n \) rounds, there always exist 4 teams with exactly one match played between them.

2. **Graph Representation:**
   Represent the teams as vertices of a graph \( G \) with 18 vertices. Each match corresponds to an edge between two vertices. After 17 rounds, the graph \( G \) is a complete graph \( K_{18} \).

3. **Subgraph Analysis:**
   We need to find the maximum \( n \) such that in the graph formed after \( n \) rounds, there always exists a subset of 4 vertices with exactly one edge among them. This is equivalent to finding a subgraph \( R \subset G \) with \( |R| = 4 \) and exactly one edge.

4. **Case for \( n = 7 \):**
   - Consider the largest group \( G \) of people, none of whom have played each other. By Zarankiewicz's Lemma, \( |G| \ge 3 \).
   - Suppose \( |G| = k \). If there exists another person not in \( G \) who played with at most \( k-2 \) people in \( G \), we win.
   - Since everyone else must play someone in \( G \) (else we could add that person to expand \( G \)), there are at most \( \frac{k \cdot 7}{k-1} \) other people.
   - Therefore, \( k + \frac{7k}{k-1} \ge 18 \), which simplifies to \( k \ge 10 \). This is absurd because nobody can have played with 9 people already.
   - Hence, \( n = 7 \) works.

5. **Case for \( n > 7 \):**
   - For \( n = 8 \), label the teams \( A_1, A_2, \ldots, A_9 \) and \( B_1, B_2, \ldots, B_9 \). On day \( i \), \( A_x \) plays \( B_{x+i} \) for \( 1 \le x \le 8 \), with indices modulo 9.
   - Check that no four people have exactly one match between them. This is easy to verify.
   - For \( n = 9 \), consider the complement of the graph for \( n = 8 \). Any four people will have played at least two matches between them.
   - For \( n \ge 10 \), adding match days to the graph for \( n = 9 \) ensures any four people have played at least two matches between them.

6. **Conclusion:**
   - For \( n \ge 8 \), it is impossible to find four teams with exactly one match between them.
   - Therefore, the maximum value of \( n \) such that there always exist 4 teams with exactly one match played between them is \( n = 7 \).

The final answer is \( n = \boxed{7} \)