Determine all integers $n \ge 2$ such that $n \mid s_n-t_n$ where $s_n$ is the sum of all the integers in the interval $[1,n]$ that are mutually prime to $n$, and $t_n$ is the sum of the remaining integers in the same interval.
1. **Define \( s_n \) and \( t_n \):**
   - \( s_n \) is the sum of all integers in the interval \([1, n]\) that are mutually prime to \( n \).
   - \( t_n \) is the sum of the remaining integers in the same interval.

2. **Express \( s_n \) using the Euler's Totient Function \( \phi(n) \):**
   - Notice that if \( \gcd(k, n) = 1 \), then \( \gcd(n - k, n) = 1 \). This implies that the set of integers \( k \) such that \( \gcd(k, n) = 1 \) can be paired with \( n - k \).
   - Therefore, we can write:
     \[
     2s_n = \sum_{\gcd(k, n) = 1} k + \sum_{\gcd(k, n) = 1} (n - k) = \sum_{\gcd(k, n) = 1} n = n \phi(n)
     \]
   - Solving for \( s_n \):
     \[
     s_n = \frac{n \phi(n)}{2}
     \]

3. **Express \( t_n \):**
   - The sum of all integers from 1 to \( n \) is:
     \[
     \sum_{k=1}^n k = \frac{n(n+1)}{2}
     \]
   - Therefore, \( t_n \) is the sum of the remaining integers:
     \[
     t_n = \frac{n(n+1)}{2} - s_n = \frac{n(n+1)}{2} - \frac{n \phi(n)}{2} = \frac{n(n+1 - \phi(n))}{2}
     \]

4. **Determine the condition for \( n \mid s_n - t_n \):**
   - We need \( n \mid s_n - t_n \):
     \[
     s_n - t_n = \frac{n \phi(n)}{2} - \frac{n(n+1 - \phi(n))}{2} = \frac{n \phi(n) - n(n+1 - \phi(n))}{2} = \frac{n \phi(n) - n^2 - n + n \phi(n)}{2} = \frac{2n \phi(n) - n^2 - n}{2}
     \]
   - Simplifying further:
     \[
     s_n - t_n = \frac{2n \phi(n) - n^2 - n}{2} = \frac{n(2 \phi(n) - n - 1)}{2}
     \]
   - For \( n \mid s_n - t_n \), we need:
     \[
     n \mid \frac{n(2 \phi(n) - n - 1)}{2}
     \]
   - This simplifies to:
     \[
     2 \phi(n) - n - 1 \text{ must be even}
     \]
   - Since \( 2 \phi(n) \) is always even, \( n + 1 \) must be even, which implies \( n \) must be odd.

Conclusion:
\[
\boxed{n \text{ is odd}}
\]