$11$ theatrical groups participated in a festival. Each day, some of the groups were scheduled to perform while the remaining groups joined the general audience. At the conclusion of the festival, each group had seen, during its days off, at least $1$ performance of every other group. At least how many days did the festival last?
To determine the minimum number of days the festival must last, we can use a combinatorial approach based on Sperner's lemma. Here is the detailed solution:

1. **Define the Problem in Terms of Sets:**
   - Let \( n = 11 \) be the number of theatrical groups.
   - Each day, a subset of these groups performs, and the rest are in the audience.
   - By the end of the festival, each group must have seen every other group perform at least once.

2. **Translate the Problem into Set Theory:**
   - Denote the groups by \( g_1, g_2, \ldots, g_{11} \).
   - Let \( G_j \) be the set of groups performing on day \( j \) for \( 1 \leq j \leq k \).
   - Define \( C_i = \{ j \mid g_i \in G_j \} \), which represents the days on which group \( g_i \) performs.

3. **Condition Translation:**
   - The condition that each group sees every other group perform translates to \( C_i \not\subseteq C_{i'} \) for any \( 1 \leq i \neq i' \leq 11 \).

4. **Use Sperner's Lemma:**
   - Sperner's lemma states that the largest antichain in the poset \( (\mathcal{P}(\{1, 2, \ldots, k\}), \subseteq) \) has size \( \binom{k}{\lfloor k/2 \rfloor} \).
   - We need \( \binom{k}{\lfloor k/2 \rfloor} \geq 11 \).

5. **Calculate the Minimum \( k \):**
   - Calculate \( \binom{k}{\lfloor k/2 \rfloor} \) for increasing values of \( k \) until the inequality is satisfied.
   - For \( k = 6 \):
     \[
     \binom{6}{3} = \frac{6!}{3!3!} = \frac{720}{6 \cdot 6} = 20
     \]
     Since \( 20 \geq 11 \), \( k = 6 \) works.

6. **Construct the Model:**
   - Choose any 11 elements from the maximal antichain of subsets of \( \{1, 2, \ldots, 6\} \) with 3 elements each.
   - Define \( G_j = \{ g_i \mid j \in C_i \} \) for \( 1 \leq j \leq 6 \).

Thus, the minimum number of days the festival must last is 6.

The final answer is \( \boxed{6} \)