There are n cups arranged on the circle. Under one of cups is hiden a coin. For every move, it is allowed to choose 4 cups and verify if the coin lies under these cups. After that, the cups are returned into its former places and the coin moves to one of two neigbor cups. What is the minimal number of moves we need in order to eventually find where the coin is?
To solve this problem, we need to determine the minimal number of moves required to find the coin under the given conditions. Let's break down the problem and the solution step by step.

1. **Understanding the Problem:**
   - We have \( n \) cups arranged in a circle.
   - A coin is hidden under one of these cups.
   - In each move, we can choose 4 cups to check if the coin is under any of them.
   - After each move, the coin moves to one of its two neighboring cups.

2. **Analyzing the Moves:**
   - For \( n = 9 \), we need to determine the minimal number of moves \( k \) required to guarantee finding the coin.
   - Let's consider the paths the coin can take in two moves. If the coin starts at position \( i \), it can move to \( i+1 \) or \( i-1 \) in the first move, and then to \( i+2 \), \( i \), or \( i-2 \) in the second move.

3. **Calculating the Paths:**
   - If \( n = 9 \) and \( k = 2 \), the first move reveals 4 cups, leaving 5 possible starting positions for the coin.
   - Each starting position can lead to 2 possible positions after the first move, resulting in \( 5 \times 2 = 10 \) possible paths.
   - In the second move, we reveal 4 more cups. Each cup can be a landing place for at least 2 paths, so at most \( 2 \times 4 = 8 \) of the 10 paths are covered.
   - This leaves at least 2 paths undetected, meaning we cannot guarantee finding the coin in 2 moves.

4. **Conclusion for \( n = 9 \):**
   - Since 2 moves are insufficient, we need at least 3 moves to ensure the coin is found.

5. **Generalizing for \( n \ge 9 \):**
   - We need to show that for \( n \ge 9 \) and \( m = \left\lceil \frac{n-3}{2} \right\rceil - 1 \), there exists a sequence of sets \( B_1, B_2, \ldots, B_m \) such that \( B_i \subset M = \{1, 2, \ldots, n\} \) and \( |B_i| = 4 \), and a sequence \( (b_1, b_2, \ldots, b_m) \) such that \( b_i \in M \setminus B_i \) for \( i = 1, 2, \ldots, m \).

6. **Inductive Proof:**
   - Base Case: For \( n = 9 \), we have shown that \( k = 3 \) is necessary.
   - Inductive Step: Assume the statement holds for \( n = k \). We need to show it holds for \( n = k + 1 \).
     - By the inductive hypothesis, there exists a sequence of sets \( B_1, B_2, \ldots, B_m \) for \( n = k \).
     - For \( n = k + 1 \), we can construct a new sequence by adding an additional set \( B_{m+1} \) and ensuring the conditions are met.

The final answer is \( \boxed{ k = \left\lceil \frac{n-3}{2} \right\rceil - 1 } \) for \( n \ge 9 \).