A circle is divided into $n$ sectors ($n \geq 3$). Each sector can be filled in with either $1$ or $0$. Choose any sector $\mathcal{C}$ occupied by $0$, change it into a $1$ and simultaneously change the symbols $x, y$ in the two sectors adjacent to $\mathcal{C}$ to their complements $1-x$, $1-y$. We repeat this process as long as there exists a zero in some sector. In the initial configuration there is a $0$ in one sector and $1$s elsewhere. For which values of $n$ can we end this process?
1. **Initial Configuration**:
   Let the numbers in the sectors be \( x_1, x_2, \dots, x_n \), and suppose that initially \( x_1 = 0 \), while \( x_i = 1 \) for \( i > 1 \). This means the initial configuration is:
   \[
   (0, 1, 1, \dots, 1)
   \]

2. **First Step**:
   After choosing the only zero available (i.e., \( x_1 \)), we change \( x_1 \) to 1 and the adjacent sectors \( x_n \) and \( x_2 \) to their complements. Thus, the new configuration is:
   \[
   (1, 0, 1, \dots, 1, 0)
   \]

3. **General Step**:
   Suppose we are in a position where \( x_n = 0 \), \( x_1 = x_2 = \dots = x_k = 0 \), \( x_{k+1} = 1 \), and \( x_{k+2} = 0 \). By choosing the zero in position \( k+2 \), we then obtain the configuration where:
   \[
   x_n = 0, \quad x_1 = x_2 = \dots = x_{k+1} = 0, \quad x_{k+2} = 1, \quad x_{k+3} = 0
   \]
   Continuing in this way, we eventually obtain the configuration where \( x_{n-2} = 1 \) and \( x_i = 0 \) otherwise.

4. **Case Analysis**:
   - **Case 1: \( n \equiv 1 \pmod{3} \)**:
     At this point, we can divide the zeros into groups of 3 and apply the procedure to the zero in the middle of each group to obtain a configuration where each sector contains a 1.

   - **Case 2: \( n \equiv 2 \pmod{3} \)**:
     We apply the procedure to the zero in sector \( n-1 \) to obtain the configuration where \( x_{n-1} = x_n = 1 \) and \( x_i = 0 \) otherwise. We can then again divide the remaining zeros into groups of 3 and apply the procedure to the zero in the middle of each group.

   - **Case 3: \( n \equiv 0 \pmod{3} \)**:
     Let \( A = x_1 + x_4 + \dots + x_{n-2} \), \( B = x_2 + x_5 + \dots + x_{n-1} \), and \( C = x_3 + x_6 + \dots + x_n \). Initially, we have:
     \[
     A = \frac{n}{3} - 1, \quad B = C = \frac{n}{3}
     \]
     In the configuration where every sector contains a 1, we have:
     \[
     A = B = C = \frac{n}{3}
     \]
     The allowed operation changes the parity of \( A \), \( B \), and \( C \). Therefore, \( A \), \( B \), and \( C \) do not all have the same parity, making it impossible to obtain a configuration where \( A = B = C \).

The final answer is \( \boxed{ n \not\equiv 0 \pmod{3} } \)