There are $n$ boyfriend-girlfriend pairs at a party. Initially all the girls sit at a round table. For the first dance, each boy invites one of the girls to dance with.After each dance, a boy takes the girl he danced with to her seat, and for the next dance he invites the girl next to her in the counterclockwise direction. For which values of $n$ can the girls be selected in such a way that in every dance at least one boy danced with his girlfriend, assuming that there are no less than $n$ dances?



1. **Labeling and Permutation Definition**:
   - Label each girl with numbers from \(1\) to \(n\).
   - We need to find a permutation \(\sigma\) of \(1, 2, \ldots, n\) such that every cyclic shift of \(\sigma\) has at least one number that matches its position.

2. **Defining \(d(i)\)**:
   - Let \(d(i) = \sigma^{-1}(i) - i\). This represents the difference between the position of \(i\) in the permutation and its original position.
   - We want each \(d(i)\) to be distinct modulo \(n\) for all \(i = 1, 2, \ldots, n\).

3. **Odd \(n\) Works**:
   - Consider the permutation where number \(i\) takes the position \(1 \leq 2i - 1 \pmod{n} \leq n\).
   - For this permutation, \(d(i) = (2i - 1) - i \equiv i - 1 \pmod{n}\).
   - Since \(i - 1\) is distinct for each \(i\), this permutation satisfies the condition.
   - Example: For \(n = 7\), the permutation is \(1, 5, 2, 6, 3, 7, 4\).

4. **Even \(n\) Does Not Work**:
   - We need to show that for even \(n\), it is impossible to find such a permutation.
   - Notice that the sum of \(d(i)\) for all \(i\) must be congruent to the sum of the first \(n\) natural numbers modulo \(n\):
     \[
     d(1) + d(2) + \cdots + d(n) \equiv 1 + 2 + \cdots + n \equiv \frac{n(n+1)}{2} \pmod{n}
     \]
   - For even \(n\), \(\frac{n(n+1)}{2}\) is not an integer multiple of \(n\), hence:
     \[
     \frac{n(n+1)}{2} \equiv \frac{n}{2} \pmod{n}
     \]
   - By definition, \(d(1) + d(2) + \cdots + d(n) \equiv (1 + 2 + \cdots + n) - (1 + 2 + \cdots + n) \equiv 0 \pmod{n}\).
   - This leads to a contradiction since \(0 \equiv \frac{n}{2} \pmod{n}\) is not possible for even \(n\).

\(\blacksquare\)

The final answer is \( \boxed{ n } \) must be odd.