 Let $n > 1$ be a natural number. We write out the fractions $\frac{1}{n}$, $\frac{2}{n}$, $\dots$ , $\dfrac{n-1}{n}$ such that they are all in their simplest form. Let the sum of the numerators be $f(n)$. For what $n>1$ is one of $f(n)$ and $f(2015n)$ odd, but the other is even?

1. **Define the function \( f(n) \)**:
   Let \( f(n) \) be the sum of the numerators of the fractions \( \frac{1}{n}, \frac{2}{n}, \dots, \frac{n-1}{n} \) that are in simplest form. This means that the numerators are those integers \( k \) such that \( 1 \leq k < n \) and \( \gcd(k, n) = 1 \).

2. **Sum of numerators in simplest form**:
   Let \( S(n) \) be the sum of the positive integers less than \( n \) that are coprime to \( n \). By properties of the Euler's totient function \( \varphi(n) \), we know that:
   \[
   S(n) = \sum_{\substack{1 \leq k < n \\ \gcd(k, n) = 1}} k
   \]
   Using the property that \( \gcd(a, n) = 1 \implies \gcd(n-a, n) = 1 \), we can pair each \( k \) with \( n - k \). Therefore, the sum of each pair is \( n \), and there are \( \frac{\varphi(n)}{2} \) such pairs. Thus:
   \[
   S(n) = \frac{n \varphi(n)}{2}
   \]

3. **Sum of numerators for \( f(n) \)**:
   Since \( f(n) \) is the sum of the numerators of the fractions in simplest form, we have:
   \[
   f(n) = S(n) = \frac{n \varphi(n)}{2}
   \]

4. **Sum of numerators for \( f(2015n) \)**:
   Similarly, for \( f(2015n) \):
   \[
   f(2015n) = \frac{2015n \varphi(2015n)}{2}
   \]

5. **Euler's totient function for \( 2015n \)**:
   Since \( 2015 = 5 \times 403 = 5 \times 13 \times 31 \), and using the multiplicative property of the totient function:
   \[
   \varphi(2015n) = \varphi(5) \varphi(13) \varphi(31) \varphi(n) = 4 \times 12 \times 30 \times \varphi(n) = 1440 \varphi(n)
   \]

6. **Calculate \( f(2015n) \)**:
   \[
   f(2015n) = \frac{2015n \cdot 1440 \varphi(n)}{2} = 2015n \cdot 720 \varphi(n)
   \]

7. **Parity condition**:
   We need to find \( n > 1 \) such that one of \( f(n) \) and \( f(2015n) \) is odd, and the other is even. This means:
   \[
   \frac{n \varphi(n)}{2} \quad \text{and} \quad 2015n \cdot 720 \varphi(n)
   \]
   must have different parities.

8. **Simplify the parity condition**:
   Since \( 2015 \) is odd and \( 720 \) is even, \( 2015n \cdot 720 \varphi(n) \) is always even. Therefore, \( \frac{n \varphi(n)}{2} \) must be odd.

9. **Condition for \( \frac{n \varphi(n)}{2} \) to be odd**:
   For \( \frac{n \varphi(n)}{2} \) to be odd, \( n \varphi(n) \) must be twice an odd number, meaning \( n \varphi(n) \) must be odd. This implies \( n \) and \( \varphi(n) \) must both be odd.

10. **Conclusion**:
    The only way for \( n \) and \( \varphi(n) \) to both be odd is if \( n \) is a power of 2. However, since \( n > 1 \), \( n \) must be an odd number greater than 1.

The final answer is \( \boxed{ n } \) must be an odd number greater than 1.