For each integer $n \ge 2$, find all integer solutions of the following system of equations:

\[x_1 = (x_2 + x_3 + x_4 + ... + x_n)^{2018}\]
\[x_2 = (x_1 + x_3 + x_4 + ... + x_n)^{2018}\]
\[\vdots\]
\[x_n = (x_1 + x_2 + x_3 + ... + x_{n - 1})^{2018}\]
1. **Define the sum \( S \):**
   Let \( S = x_1 + x_2 + \cdots + x_n \). Then, each equation in the system can be rewritten as:
   \[
   x_i = (S - x_i)^{2018} \quad \text{for } i = 1, 2, \ldots, n.
   \]

2. **Analyze the equation \( x_i = (S - x_i)^{2018} \):**
   Since \( x_i \) is an integer, \( (S - x_i)^{2018} \) must also be an integer. This implies that \( S - x_i \) must be an integer as well.

3. **Consider the possible values of \( S - x_i \):**
   Since \( (S - x_i)^{2018} \) is a perfect 2018th power, \( S - x_i \) must be an integer. Let \( y_i = S - x_i \). Then, the equation becomes:
   \[
   x_i = y_i^{2018}.
   \]
   Substituting \( y_i = S - x_i \) into the equation, we get:
   \[
   x_i = (S - x_i)^{2018}.
   \]

4. **Examine the implications of \( x_i = (S - x_i)^{2018} \):**
   If \( x_i = 0 \), then \( (S - x_i)^{2018} = 0 \), which implies \( S - x_i = 0 \) or \( S = x_i \). This would mean \( S = 0 \) and all \( x_i = 0 \).

5. **Consider non-zero solutions:**
   If \( x_i \neq 0 \), then \( x_i \) must be a perfect 2018th power. Let \( x_i = k_i^{2018} \) for some integer \( k_i \). Then:
   \[
   k_i^{2018} = (S - k_i^{2018})^{2018}.
   \]
   Taking the 2018th root of both sides, we get:
   \[
   k_i = S - k_i^{2018}.
   \]
   This implies:
   \[
   S = k_i + k_i^{2018}.
   \]

6. **Check consistency across all \( x_i \):**
   Since \( S \) is the same for all \( i \), we must have:
   \[
   S = k_1 + k_1^{2018} = k_2 + k_2^{2018} = \cdots = k_n + k_n^{2018}.
   \]
   This implies that all \( k_i \) must be equal. Let \( k = k_i \) for all \( i \). Then:
   \[
   S = k + k^{2018}.
   \]

7. **Solve for \( k \):**
   Since \( S = n \cdot k^{2018} \) (because \( x_i = k^{2018} \) for all \( i \)), we have:
   \[
   n \cdot k^{2018} = k + k^{2018}.
   \]
   Simplifying, we get:
   \[
   (n-1)k^{2018} = k.
   \]
   If \( k = 0 \), then \( x_i = 0 \) for all \( i \), which is a solution.

8. **Consider \( k \neq 0 \):**
   If \( k \neq 0 \), we can divide both sides by \( k \):
   \[
   (n-1)k^{2017} = 1.
   \]
   Since \( k \) is an integer, \( k^{2017} \) must be 1, which implies \( k = 1 \) or \( k = -1 \). However, \( k = -1 \) is not possible because \( k^{2018} \) would be positive. Thus, \( k = 1 \).

9. **Verify the solution \( k = 1 \):**
   If \( k = 1 \), then \( x_i = 1 \) for all \( i \), and \( S = n \cdot 1 = n \). This satisfies the original equations.

Conclusion:
The integer solutions to the system are \( x_i = 0 \) for all \( i \) or \( x_i = 1 \) for all \( i \).

The final answer is \( \boxed{ x_i = 0 } \) for all \( i \) or \( x_i = 1 \) for all \( i \).