Let $n \geq 2$ be an integer. Define  $x_i =1$ or $-1$ for every $i=1,2,3,\cdots, n$.

Call an operation [i]adhesion[/i], if it changes the string $(x_1,x_2,\cdots,x_n)$ to $(x_1x_2, x_2x_3, \cdots ,x_{n-1}x_n, x_nx_1)$ .

 Find all integers $n \geq 2$ such that the string  $(x_1,x_2,\cdots, x_n)$ changes to $(1,1,\cdots,1)$ after finitely [i]adhesion[/i] operations.
1. **Define the problem and the operation:**
   - We are given a sequence \( (x_1, x_2, \ldots, x_n) \) where each \( x_i \) is either \( 1 \) or \( -1 \).
   - The operation called *adhesion* transforms the sequence \( (x_1, x_2, \ldots, x_n) \) into \( (x_1 x_2, x_2 x_3, \ldots, x_{n-1} x_n, x_n x_1) \).

2. **Consider the infinite sequence:**
   - Extend the sequence \( (x_1, x_2, \ldots, x_n) \) periodically to an infinite sequence \( (x_1, x_2, \ldots, x_n, x_1, x_2, \ldots) \).
   - The *adhesion* operation on this infinite sequence results in a new sequence \( (y_1, y_2, \ldots) \) where \( y_k = x_k x_{k+1} \).

3. **Properties of the sequence:**
   - Since \( x_i^2 = 1 \) for all \( i \), the sequence remains unchanged if we "erase" the squares in them.

4. **Define multiple adhesion operations:**
   - Let \( \text{adhesion}_k \) denote the operation *adhesion* applied \( 2^k \) times.
   - Note that \( \text{adhesion}_0 \) is the original *adhesion* operation, and \( \text{adhesion}_{k+1} \) is twice \( \text{adhesion}_k \).

5. **Condition for the sequence to become all ones:**
   - If for some \( n \), every sequence \( (x_1, x_2, \ldots) \) turns into \( (1, 1, \ldots) \) after a finite number of *adhesion* operations, there exists an integer \( A \) such that applying \( \text{adhesion} \) \( A \) times results in \( (1, 1, \ldots) \).

6. **Manipulation of variables:**
   - If after erasing the squares there exist variables, we can manipulate them to make at least one of them \( -1 \). Therefore, \( n \) will work only if the sequence \( (x_i) \) turns into \( (1, 1, \ldots) \) after \( A \) operations.

7. **Prove the form of the sequence after multiple adhesions:**
   - We will prove that \( \text{adhesion}_k \) of \( (x_1, x_2, \ldots) \) is \( (x_1 x_{2^k+1}, x_2 x_{2^k+2}, \ldots) \).
   - This is true for \( k = 0 \). Suppose it is true for some \( k \), then \( \text{adhesion}_{k+1} \) of \( \text{adhesion}_k \) of \( (x_1, x_2, \ldots) \) equals \( \text{adhesion}_k \) of \( (x_1 x_{2^k+1}, x_2 x_{2^k+2}, \ldots) \).
   - This results in \( ((x_1 x_{2^k+1})(x_{2^k+1} x_{2^{k+1}+1}), (x_1 x_{2^k+2})(x_{2^k+2} x_{2^{k+1}+2}), \ldots) \) which equals \( (x_1 x_{2^{k+1}+1}, x_2 x_{2^{k+1}+2}, \ldots) \).
   - Thus, it is true for \( k+1 \).

8. **Conclusion for large \( A \):**
   - For sufficiently large \( A \), \( \text{adhesion}_A \) of \( (x_1, x_2, \ldots) \) is \( (1, 1, \ldots) \) but also \( (x_1 x_{2^A+1}, x_2 x_{2^A+2}, \ldots) \).
   - Therefore, \( r \equiv 2^A + r \mod n \) implies \( n \mid 2^A \), so \( n \) must be a power of 2.

9. **Verification:**
   - If \( n = 2^m \), then \( \text{adhesion}_m \) (or equivalently \( 2^m \) times *adhesion*) results in \( (1, 1, \ldots) \).

The final answer is \( \boxed{ n } \) must be a power of 2.