Let the total number of performance days be $n$, and the set of days on which the $i$-th troupe does not perform be $A_{i}$ $(i=1,2, \cdots, 11)$. Then $A_{1}, A_{2}, \cdots, A_{11}$ are all subsets of $X=\{1, 2, \cdots, n\}$.

Since each troupe has seen the performances of all other troupes, $A_{i}, A_{j} (1 \leqslant i<j \leqslant 11)$ do not contain each other (the day when the $i$-th troupe watches the $j$-th troupe's performance belongs to $A_{i}$ but not to $A_{j}$). By Sperner's theorem,
$$
11 \leqslant C_{n}^{\left[\frac{n}{2}\right]} \text {. }
$$
$C_{n}^{\left[\frac{1}{2}\right]}$ increases with $n$, $C_{5}^{2}=10, C_{6}^{3}=20$, so $n \geqslant 6$. That is, there must be at least 6 performance days.
For example, among 211 theater troupes, some troupes perform each day while others watch (those who perform cannot watch). If each troupe has seen performances by 10 other troupes, how many days are needed for the performances at least?



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