Find all pairs of positive integers $(m,n)$ for which there exist relatively prime integers $a$ and $b$ greater than $1$ such that
$$\frac{a^m+b^m}{a^n+b^n}$$
is an integer.
To find all pairs of positive integers \((m, n)\) for which there exist relatively prime integers \(a\) and \(b\) greater than 1 such that 
\[ \frac{a^m + b^m}{a^n + b^n} \]
is an integer, we proceed as follows:

1. **Case \(m = n\)**:
   - If \(m = n\), then \(\frac{a^m + b^m}{a^n + b^n} = \frac{a^n + b^n}{a^n + b^n} = 1\), which is an integer. Therefore, \((m, n) = (k, k)\) for any positive integer \(k\) is a solution.

2. **Case \(m \neq n\)**:
   - Assume \(m > n\). Without loss of generality, let \(b > a\). We need \(a^n + b^n \mid a^m + b^m\).
   - This implies \(a^n + b^n \mid b^m - (-a)^m\). Since \(a\) and \(b\) are relatively prime, \(\gcd(a^n + b^n, b) = 1\). Therefore, \(a^n + b^n \mid b^{m-n} - a^{m-n}\).

3. **Case \(m - n = n\)**:
   - If \(m = 2n\), then \(a^n + b^n \mid b^n(b^{n-1} - a^{n-1})\). Since \(\gcd(a^n + b^n, b^n) = 1\), we have \(a^n + b^n \mid b^{n-1} - a^{n-1}\).
   - However, since \(a, b > 1\), \(a^n > 1\) and \(b^n > b^{n-1}\), thus \(b^{n-1} + 1 < a^n + b^n\), which is a contradiction. Therefore, \(m \neq 2n\).

4. **Case \(m - n < n\)**:
   - From \(a^n + b^n \mid b^{m-n} - a^{m-n}\), we have \(0 < b^{m-n} - a^{m-n} < b^n < a^n + b^n\), which is a contradiction. Therefore, \(m - n \geq n\).

5. **Case \(m - n > n\)**:
   - We still have \(a^n + b^n \mid b^{m-n} - a^{m-n}\). Applying the Euclidean algorithm, we get \(a^n + b^n \mid b^{m-2n} + a^{m-2n}\).
   - If \(m - 2n < n\), we reach a contradiction. Thus, either \(m - 2n = n\) or \(m = 3n\), which is a solution, or \(m > 3n\).

6. **General Case**:
   - If \(m > 3n\), we continue applying the Euclidean algorithm: \(a^n + b^n \mid b^{m-3n} - a^{m-3n}\).
   - In general, if \((2k-1)n \leq m < 2kn\) for some positive integer \(k\), then \(a^n + b^n \mid b^{m-(2k-2)n} + a^{m-(2k-2)n}\).
   - By induction, unless \(m - (2k-2)n = n\) or \(m = (2k-1)n\), we won't have a solution.

Therefore, the pairs \((m, n)\) that satisfy the condition are \((m, n) = ((2k-1)n, n)\) where \(n\) and \(k\) are positive integers.

The final answer is \( \boxed{ (m, n) = ((2k-1)n, n) } \) where \(n\) and \(k\) are positive integers.