Given an integer $n\geq 2$, determine the maximum value the sum $x_1+\cdots+x_n$ may achieve, as the $x_i$ run through the positive integers, subject to $x_1\leq x_2\leq \cdots \leq x_n$ and $x_1+\cdots+x_n=x_1 x_2\cdots x_n$.
To determine the maximum value of the sum \( x_1 + x_2 + \cdots + x_n \) subject to the conditions \( x_1 \leq x_2 \leq \cdots \leq x_n \) and \( x_1 + x_2 + \cdots + x_n = x_1 x_2 \cdots x_n \), we can proceed as follows:

1. **Define the sum and product:**
   Let \( S_i = \sum_{j=1}^i x_j \) and \( P_i = \prod_{j=1}^i x_j \). We need to find the maximum \( S_n \) such that \( S_n = P_n \).

2. **Express \( x_n \) in terms of \( S_{n-1} \) and \( P_{n-1} \):**
   Since \( S_n = P_n \), we can write:
   \[
   x_n = \frac{S_{n-1}}{P_{n-1} - 1}
   \]
   because \( S_n = S_{n-1} + x_n \) and \( P_n = P_{n-1} \cdot x_n \).

3. **Ensure \( x_n \geq 2 \):**
   Given \( x_n \geq 2 \), we have:
   \[
   S_{n-1} \geq 2(P_{n-1} - 1) \geq P_{n-1}
   \]
   This implies \( S_{n-1} \geq P_{n-1} \).

4. **Inductive step:**
   For \( x_i \geq 2 \), we need \( S_{i-1} \geq P_{i-1} \). If \( x_1 = 2 \), then for \( n = 2 \), \( x_2 = 2 \). For other solutions, we have \( x_1 = x_2 = \cdots = x_k = 1 \) and \( 2 \leq x_{k+1} \leq x_{k+2} \leq \cdots \leq x_n \).

5. **Factorization approach:**
   For a given \( S = S_n = P_n \), we can factorize \( S \) as \( S = (y_1 + 1)(y_2 + 1) \cdots (y_l + 1) \) where \( 1 \leq y_1 \leq y_2 \leq \cdots \leq y_l \). We find \( k = S - (y_1 + 1) - (y_2 + 1) - \cdots - (y_l + 1) \) and \( n = k + l = S - y_1 - y_2 - \cdots - y_l \).

6. **Prime consideration:**
   \( S \) is not prime. If \( l = 2 \), then \( S = (a + 1)(b + 1) \) and \( n = (a + 1)(b + 1) - a - b = ab + 1 \). Thus, \( S \leq 2n \).

7. **General case:**
   If \( l \geq 3 \), then \( S = (y_1 + 1)(y_2 + 1) \cdots (y_l + 1) \) and \( n = S - y_1 - y_2 - \cdots - y_l \). We have:
   \[
   S = n \frac{1}{1 - x}, \quad x = \frac{\sum_i y_i}{\prod_i (1 + y_i)} \leq \frac{1}{2} \implies S \leq 2n
   \]
   \( S = 2n \) only if \( x_n = n, x_{n-1} = 2, x_i = 1 \) for \( i < n - 1 \).

The final answer is \( \boxed{ S = 2n } \).