For each natural number $n\geq 2$, solve the following system of equations in the integers $x_1, x_2, ..., x_n$:
$$(n^2-n)x_i+\left(\prod_{j\neq i}x_j\right)S=n^3-n^2,\qquad \forall 1\le i\le n$$
where
$$S=x_1^2+x_2^2+\dots+x_n^2.$$
Given the system of equations for each natural number \( n \geq 2 \):

\[
(n^2 - n)x_i + \left(\prod_{j \neq i} x_j\right) S = n^3 - n^2, \quad \forall 1 \leq i \leq n
\]

where 

\[
S = x_1^2 + x_2^2 + \dots + x_n^2
\]

we need to solve for the integers \( x_1, x_2, \ldots, x_n \).

1. **Assume \(\prod_{i=1}^n x_i = P = 0\)**:
   - If \( P = 0 \), then there exists some \( k \in \{1, 2, \ldots, n\} \) such that \( x_k = 0 \).
   - For \( i \neq k \), the equation becomes:
     \[
     (n^2 - n)x_i = n^3 - n^2 \implies x_i = n
     \]
   - Substituting \( x_i = n \) for all \( i \neq k \) into the equation for \( x_k \):
     \[
     (n^2 - n) \cdot 0 + n^{n-1} \cdot S = n^3 - n^2
     \]
     where \( S = n^2(n-1) \).
   - This implies:
     \[
     n^{n-1} = 1 \implies n = 1
     \]
     which contradicts \( n \geq 2 \). Hence, \( P \neq 0 \).

2. **Assume \( x_i \neq 0 \) for all \( i \in \{1, 2, \ldots, n\} \)**:
   - Multiply each side of the equation by \( x_i \):
     \[
     (n^2 - n)x_i^2 + PS = (n^3 - n^2)x_i
     \]
     which simplifies to:
     \[
     (n^2 - n)x_i(n - x_i) = PS
     \]

3. **Consider \( i \neq j \)**:
   - From the above equation, we have:
     \[
     (n^2 - n)x_i(n - x_i) = (n^2 - n)x_j(n - x_j)
     \]
     which simplifies to:
     \[
     (x_i - x_j)(x_i + x_j - n) = 0
     \]
     Therefore, \( x_i = x_j \) or \( x_i + x_j = n \).

4. **Set \( j = 1 \)**:
   - We obtain \( x_i \in \{x_1, n - x_1\} \) for all \( i \in \{1, 2, \ldots, n\} \).
   - Let \( k = |\{x_i = x_1 | i \in \{1, 2, \ldots, n\}\}| \).

5. **Consider \( k = n \)**:
   - If \( k = n \), then all \( x_i = x_1 \):
     \[
     (n^2 - n)x_1(n - x_1) = x_1^n \cdot nx_1^2
     \]
     which simplifies to:
     \[
     x_1^{n+1} + (n-1)x_1 = n(n-1)
     \]
   - If \( x_1 \geq 2 \), then:
     \[
     n(n-1) \geq 2^{n+1} + 2(n-1)
     \]
     which contradicts \( 2^{n+1} > n(n-1) \) for \( n \geq 3 \). Hence, \( x_1 = 1 \) is the only possibility, leading to \( n = 2 \).

6. **Consider \( k < n \)**:
   - By setting \( k \in \{1, n-1\} \), we find:
     \[
     x_1 = n - k
     \]
   - For \( k = 1 \), \( x_1 = n-1 \) and \( x_i = 1 \) for all \( i \geq 2 \).
   - For \( k = n-1 \), \( x_i = 1 \) for all \( i \leq n-1 \) and \( x_n = n-1 \).

Thus, the solutions of the system of equations are \( x_i = 1 \) and \( x_j = n-1 \) for all \( j \neq i \).

The final answer is \( \boxed{ x_i = 1 } \) and \( x_j = n-1 \) for all \( j \neq i \).