5.69 11 theater troupes participate in a performance, with some of the troupes scheduled to perform each day, while the rest join the general audience. By the end of the performance, each troupe, apart from their own performance days, must have watched at least one performance by each of the other troupes. How many days at minimum must the performance be scheduled for?
[Solution] Let $N=\{1,2, \cdots, n\}$ be the set of performance dates.
Each theater group can be represented by a subset $A \subset N$: Theater group $A$ performs on day $x$ if and only if $x \in A$.

For two theater groups $A$ and $B$ to watch each other's performances at least once, there must be at least one day that belongs to $B$ but not to $A$, and another day that belongs to $A$ but not to $B$. Therefore, $A \subseteq B$ or $B \subseteq A$ cannot hold, and in particular, $A \neq B$. Thus, different theater groups must have different subsets of performance dates.
We first prove that $n=6$ is sufficient.
Indeed, one possible set of performance dates for 11 theater groups is:
$$
\begin{array}{l}
\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\}, \\
\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\},\{6\} .
\end{array}
$$

This is exactly 11 subsets of the date set $N=\{1,2,3,4,5,6\}$, and no two subsets are contained in each other.

Next, we prove that $n=5$ is not sufficient. Because each theater group's set of performance dates $A \subset\{1,2,3,4,5\}$, and clearly $1 \leqslant|A| \leqslant 4$ (where $|A|$ denotes the number of elements in set $A$).

Define a chain as a sequence $A_{1} \subset A_{2} \subset A_{3} \subset A_{4}$, where $\left|A_{i}\right|=i, 1 \leqslant i \leqslant 4$. Then an element of $A_{1}$ can be any one of the numbers $1,2,3,4,5$. $A_{2}$ contains 2 elements, one of which is an element of $A_{1}$, $A_{3}$ contains 3 elements, 2 of which are elements of $A_{2}$, and similarly, $A_{4}$ contains 4 elements, 3 of which are elements of $A_{3}$. Therefore, the number of such chains is 120. Each subset with 1 or 4 elements appears in $4 \times 3 \times 2=24$ chains, and each subset with 2 or 3 elements appears in $2 \times 3 \times 2=12$ chains (for example, $A_{2}$ contains two numbers, $A_{1}$ contains one of these two numbers, $A_{3}$ contains one of the other three numbers, and $A_{4}$ contains one of the remaining two numbers). Since we have 11 theater groups, each group's set of performance dates appears 12 or 24 times in the 120 chains, so the 11 sets of performance dates appear at least $11 \times 12=132$ times in total. By the pigeonhole principle, at least two sets of performance dates must appear in the same chain, denoted as $A$ and $B$, which contradicts the fact that $A$ and $B$ do not contain each other.
Therefore, such an arrangement requires at least 6 days.