Example 1: $n$ teams are to hold a home-and-away double round-robin match (each pair of teams plays two matches, each having one home match), and each team can play multiple away matches within a week (from Sunday to Saturday). However, if a team has a home match in a certain week, no away matches for that team can be scheduled within the same week. If all matches can be completed within 4 weeks, find the maximum value of $n$.
(Note: A match held at Team A's venue is called Team A's home match and Team B's away match.)
Solution (1) As shown in the table below: An asterisk (*) in the table indicates that the team has a home game in that week and cannot travel. It is easy to verify that, according to the schedule in the table, the six teams can complete the matches in four weeks.
\begin{tabular}{|c|c|c|c|c|}
\hline \begin{tabular}{l} 
Team \\
\end{tabular} & \begin{tabular}{l} 
1st
\end{tabular} & \begin{tabular}{l} 
2nd
\end{tabular} & \begin{tabular}{l}
$\stackrel{\text { 3rd }}{\overline{=}}$
\end{tabular} & \begin{tabular}{l} 
4th
\end{tabular} \\
\hline 1 & * & * & & \\
\hline 2 & * & & * & \\
\hline 3 & * & & & * \\
\hline 4 & & * & * & \\
\hline 5 & & * & & * \\
\hline 6 & & & * & * \\
\hline
\end{tabular}
(2) The following proves that seven teams cannot complete the matches in four weeks. Let $S_{i}(i=1,2,3,4,5,6,7)$ represent the set of weeks in which team $i$ has home games. Assuming the matches can be completed in four weeks, then $S_{i}$ is a non-empty proper subset of $\{1,2,3,4\}$.

On one hand, since a team cannot have an away game in a week when it has a home game, none of the sets $S_{i}(i=1,2,3,4,5,6,7)$ is a subset of another.

On the other hand, let $A=\{\{1\},\{1,2\},\{1,2,3\}\}, B=\{\{2\},\{2,3\},\{2,3,4\}\}, C=\{\{3\},\{1,3\},\{1,3,4\}\}, D=\{\{4\},\{1,4\},\{1,2,4\}\}, E=\{\{2,4\}\}, F=\{\{3,4\}\}$. By the pigeonhole principle, there must exist $i, j \in\{1,2,3,4,5\}, i \neq j$, such that $S_{i}$ and $S_{j}$ belong to the same set $A$ or $B$ or $C$ or $D$ or $E$ or $F$, thus either $S_{i} \subseteq S_{j}$ or $S_{i} \supseteq S_{j}$ must occur.

In conclusion, the maximum value of $n$ is 6.