Seven, $n$ teams are to hold a home-and-away double round-robin match (each 2 teams play 2 matches, each having 1 home match), 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 in that week. If all matches can be completed within four weeks, find the maximum value of $n$.

Note: A match held at $A$'s venue between teams $A$ and $B$ is called $A$'s home match and $B$'s away match.
(Supplied by Zonghu Biao)
(1) As shown in Table 1. The symbol “*” 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 competition in four weeks.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Team & Week 1 & Week 2 & Week 3 & Week 4 \\
\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 competition in four weeks. Let $S_{i}(i=1,2, \cdots, 7)$ represent the set of weeks in which the $i$-th team has home games. Assuming the competition can be completed within 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 $S_{i}(i=1,2, \cdots, 7)$ is a subset of another.
On the other hand, let
$$
\begin{array}{l}
A=\{\{1\},\{1,2\},\{1,2,3\}\}, \\
B=\{\{2\},\{2,3\},\{2,3,4\}\}, \\
C=\{\{3\},\{1,3\},\{1,3,4\}\}, \\
D=\{\{1\},\{1,1\},\{1,2,1\}\}, \\
E=\{\{2,4\}\}, F=\{\{3,4\}\} .
\end{array}
$$

By the pigeonhole principle, there must exist $i, j, i \neq j, i, j \in\{1,2,3,4,5,6,7\}$ such that $S_{i}$ and $S_{j}$ belong to the same set $A$ or $B$ or $C$ or $D$ or $E$ or $F$, and thus $S_{i} \subseteq S_{j}$ or $S_{i} \supseteq S_{j}$ must occur.
Therefore, the maximum value of $n$ is 6.