Given a positive integer $n \geq 2$. Solve the following system of equations:
$
\begin{cases}
\ x_1|x_1| &= x_2|x_2| + (x_1-1)|x_1-1| \\
\ x_2|x_2| &= x_3|x_3| + (x_2-1)|x_2-1| \\
&\dots \\
\ x_n|x_n| &= x_1|x_1| + (x_n-1)|x_n-1|. \\
\end{cases}
$
1. **Given System of Equations:**
   \[
   \begin{cases}
   x_1|x_1| = x_2|x_2| + (x_1-1)|x_1-1| \\
   x_2|x_2| = x_3|x_3| + (x_2-1)|x_2-1| \\
   \vdots \\
   x_n|x_n| = x_1|x_1| + (x_n-1)|x_n-1|
   \end{cases}
   \]

2. **Summing All Equations:**
   \[
   \sum_{i=1}^n x_i|x_i| = \sum_{i=1}^n \left( x_{i+1}|x_{i+1}| + (x_i-1)|x_i-1| \right)
   \]
   Note that \(x_{n+1} = x_1\) due to the cyclic nature of the equations.

3. **Simplifying the Summation:**
   \[
   \sum_{i=1}^n x_i|x_i| = \sum_{i=1}^n x_{i+1}|x_{i+1}| + \sum_{i=1}^n (x_i-1)|x_i-1|
   \]
   Since \(\sum_{i=1}^n x_{i+1}|x_{i+1}|\) is just a reordering of \(\sum_{i=1}^n x_i|x_i|\), we have:
   \[
   \sum_{i=1}^n x_i|x_i| = \sum_{i=1}^n x_i|x_i| + \sum_{i=1}^n (x_i-1)|x_i-1|
   \]

4. **Isolating the Additional Term:**
   \[
   0 = \sum_{i=1}^n (x_i-1)|x_i-1|
   \]

5. **Analyzing the Term \((x_i-1)|x_i-1|\):**
   - If \(x_i > 1\), then \((x_i-1)|x_i-1| = (x_i-1)(x_i-1) > 0\).
   - If \(x_i < 1\), then \((x_i-1)|x_i-1| = (x_i-1)(1-x_i) < 0\).
   - If \(x_i = 1\), then \((x_i-1)|x_i-1| = 0\).

6. **Considering \(x_i > 1\) for Some \(i\):**
   - If \(x_i > 1\), then \((x_i-1)|x_i-1| > 0\).
   - This implies \(\sum_{i=1}^n (x_i-1)|x_i-1| > 0\), which contradicts the equation \(0 = \sum_{i=1}^n (x_i-1)|x_i-1|\).

7. **Considering \(x_i < 1\) for Some \(i\):**
   - If \(x_i < 1\), then \((x_i-1)|x_i-1| < 0\).
   - This implies \(\sum_{i=1}^n (x_i-1)|x_i-1| < 0\), which also contradicts the equation \(0 = \sum_{i=1}^n (x_i-1)|x_i-1|\).

8. **Conclusion:**
   - The only possibility left is \(x_i = 1\) for all \(i\), which makes \((x_i-1)|x_i-1| = 0\) for all \(i\).
   - This satisfies the original system of equations.

\(\blacksquare\)

The final answer is \( \boxed{ x_i = 1 \quad \forall i } \).