$7 n$ teams are to hold a double round-robin match (each pair of teams plays two matches, one home and one away match). Each team can play multiple matches within a week (from Sunday to Saturday). If a team has $1:1$ matches in a week, no away matches can be scheduled for this team in the same week. If all matches can be completed within 4 weeks, find the maximum value of $n$. (Proposed by Zonghu)

Note: The match held at the home venue of Team A and the away venue of Team B is called the home match of Team A and the away match of Team B.
(1) As shown in the table: The table has “*”, indicating that the team has a match in that week. It is easy to verify that, according to the schedule in the table, 6 teams can complete the matches in that week.
(2) The following proves that 7 teams cannot complete the matches in that week. Let $S_{i}(i=1, 2, 3, 4, 5, 6, 7)$ represent the set of weeks in which the $i$-th team plays home matches. Assuming 4 teams can complete the matches, then $S$ is the set of all matches of that team. Therefore, $S_{i}(i=1, 2, 3, 4, 5, 6, 7)$ are all non-empty sets.

Each set is a subset of the following sets:
$$
\begin{array}{l}
A=\{\{1\},\{1,2\},\{1,2,3\}\} \\
B=\{\{2\},\{2,3\},\{2,3,4\}\} \\
C=\{\{3\},\{1,3\},\{1,3,4\}\} \\
D=\{\{4\},\{1,4\},\{1,2,4\}\} \\
E=\{\{2,4\}\} \\
F=\{\{3,4\}\}
\end{array}
$$
Since $S_{i}$ is a subset of one of the sets $A$, $B$, $C$, $D$, $E$, or $F$, there must be $S_{i} \subseteq S$ or $S_{i} \neq S$.
In conclusion, the maximum value of $n$ is 6.