At the entrance to a cave is a rotating round table. On top of the table are $n$ identical barrels, evenly spaced along its circumference. Inside each barrel is a herring either with its head up or its head down. In a move, Ali Baba chooses from $1$ to $n$ of the barrels and turns them upside down. Then the table spins around. When it stops, it is impossible to tell which barrels have been turned over. The cave will open if the heads of the herrings in all $n$ barrels are up or are all down. Determine all values of $n$ for which Ali Baba can open the cave in a finite number of moves.

[i](11 points)[/i]
To determine all values of \( n \) for which Ali Baba can open the cave in a finite number of moves, we need to analyze the problem in detail. The problem can be broken down into several cases based on the value of \( n \).

1. **Case 1: \( n \neq 2^k \) for \( k = 0, 1, 2, \ldots \)**

   - If \( n \) is not a power of 2, we need to show that Ali Baba cannot guarantee opening the cave in a finite number of moves.
   - Assume \( n \) is odd. We can place the digits "0" and "1" in such a way that after Ali Baba's first move, there are still both "0"s and "1"s present. This means that the number of "0"s and "1"s is not equal, as their sum is odd.
   - If Ali Baba selects \( k \) positions, he wins if and only if these \( k \) positions coincide with either all "0"s or all "1"s. However, since the number of "0"s and "1"s is not equal, at least one of these numbers differs from \( k \). By moving such a digit to one of the selected \( k \) positions, we prevent Ali Baba from winning in this round.
   - If \( n \) is even but has an odd factor \( m \), we can reduce this case to the previous one by marking \( m \) equidistant positions and ignoring the others. Using the same method as above, we can prevent Ali Baba from making all digits in these \( m \) positions equal.

2. **Case 2: \( n = 2^k \) for \( k = 0, 1, 2, \ldots \)**

   - We need to construct an algorithm for Ali Baba to win when \( n \) is a power of 2. We will use induction to build this algorithm.
   - **Base Case:** For \( k = 1 \), \( n = 2 \). This is trivial as Ali Baba can simply flip both barrels to make them the same.
   - **Inductive Step:** Assume Ali Baba has an algorithm \( A_m \) for \( m \) barrels. We need to construct \( A_{2m} \) for \( 2m \) barrels.
     - Split the \( 2m \) positions into \( m \) pairs of opposite positions. Establish a one-to-one correspondence between these pairs and the positions for \( n = m \).
     - **Special Case (a):** If in each pair the digits are the same, apply \( A_m \) to the pairs. If \( A_m \) unlocks the cave for \( n = m \), it will also unlock the cave for \( n = 2m \).
     - **Special Case (b):** Consider the parity of the sum in each pair. If all parities are the same, apply \( A_m \). If the cave is unlocked, all parities were odd. Changing both digits in a pair does not change the parity, so all parities remain odd. Change exactly one digit in each pair (operation \( D \)). Now all parities become even. Apply \( A_m \) again to unlock the cave. Call this algorithm \( B \).
     - **General Case (c):** Apply \( A_m \) in a different way to make all parities equal. Change only one digit in each pair (algorithm \( C \)). This guarantees that at some step all parities will coincide. After each step of applying \( C \), apply \( B \) and then \( D \). If after \( C \) the parities coincide, \( B \) unlocks the cave. Otherwise, \( BD \) does not change parities, so parities after \( C \) and \( CBD \) are the same. Since \( \underbrace{C...C}_{j \text{ times}} \) makes parities equal, \( \underbrace{CBD...CBD}_{j \text{ times}} \) also makes parities equal. Therefore, \( \underbrace{CBD...CBD}_{j-1 \text{ times}}CB \) unlocks the cave.

The final answer is \( \boxed{ n = 2^k } \) for \( k = 0, 1, 2, \ldots \)