A parliament has $n$ senators. The senators form 10 parties and 10 committees, such that any senator belongs to exactly one party and one committee. Find the least possible $n$ for which it is possible to label the parties and the committees with numbers from 1 to 10, such that there are at least 11 senators for which the numbers of the corresponding party and committee are equal.
1. Let \( n_0 \) be the number of senators for which the party number equals the committee number, i.e., \((\#\text{party}) = (\#\text{committee})\).
2. Let \( n_1 \) be the number of senators for which \((\#\text{party}) - (\#\text{committee}) \equiv 1 \pmod{10}\).
3. Similarly, let \( n_2 \) be the number of senators for which \((\#\text{party}) - (\#\text{committee}) \equiv 2 \pmod{10}\), and so on up to \( n_9 \).

Thus, we have:
\[ n = n_0 + n_1 + n_2 + \cdots + n_9 \]

4. We need to find the smallest \( n \) such that \( n_0 \geq 11 \). By the pigeonhole principle, if we distribute \( n \) senators into 10 categories (corresponding to the 10 possible values of \((\#\text{party}) - (\#\text{committee}) \pmod{10}\)), at least one category must contain at least \( \lceil \frac{n}{10} \rceil \) senators.

5. To ensure that \( n_0 \geq 11 \), we need:
\[ \lceil \frac{n}{10} \rceil \geq 11 \]

6. Solving for \( n \):
\[ \frac{n}{10} \geq 11 \]
\[ n \geq 110 \]

7. Therefore, the smallest \( n \) that satisfies this condition is \( n = 110 \).

The final answer is \( \boxed{110} \).