Find all positive integers that can be written in the following way $\frac{m^2 + 20mn + n^2}{m^3 + n^3}$
Also, $m,n$ are relatively prime positive integers.
1. We start with the given expression:
   \[
   \frac{m^2 + 20mn + n^2}{m^3 + n^3}
   \]
   We can rewrite the numerator and the denominator in a more convenient form. Notice that:
   \[
   m^3 + n^3 = (m+n)(m^2 - mn + n^2)
   \]
   and
   \[
   m^2 + 20mn + n^2 = (m+n)^2 + 18mn
   \]
   Therefore, the given expression becomes:
   \[
   \frac{(m+n)^2 + 18mn}{(m+n)(m^2 - mn + n^2)}
   \]

2. Simplify the fraction:
   \[
   \frac{(m+n)^2 + 18mn}{(m+n)(m^2 - mn + n^2)} = \frac{(m+n)^2 + 18mn}{(m+n)(m^2 - mn + n^2)}
   \]
   Since \(m\) and \(n\) are relatively prime, \(\gcd(m+n, mn) = 1\). This implies that \((m+n)\) must divide \(18\) because \((m+n)\) divides the numerator \((m+n)^2 + 18mn\).

3. Determine the possible values of \(m+n\):
   \[
   m+n \mid 18 \implies m+n = 1, 2, 3, 6, 9, 18
   \]

4. Check each possible value of \(m+n\) to find valid pairs \((m, n)\):

   - For \(m+n = 1\):
     \[
     \text{No positive integer solutions since } m \text{ and } n \text{ are positive integers.}
     \]

   - For \(m+n = 2\):
     \[
     \text{No positive integer solutions since } m \text{ and } n \text{ are relatively prime.}
     \]

   - For \(m+n = 3\):
     \[
     (m, n) = (2, 1) \text{ or } (1, 2)
     \]

   - For \(m+n = 6\):
     \[
     (m, n) = (5, 1) \text{ or } (1, 5)
     \]

   - For \(m+n = 9\):
     \[
     \text{No relatively prime positive integer solutions.}
     \]

   - For \(m+n = 18\):
     \[
     \text{No relatively prime positive integer solutions.}
     \]

5. Verify the solutions:
   - For \((m, n) = (2, 1)\):
     \[
     \frac{2^2 + 20 \cdot 2 \cdot 1 + 1^2}{2^3 + 1^3} = \frac{4 + 40 + 1}{8 + 1} = \frac{45}{9} = 5
     \]
   - For \((m, n) = (1, 2)\):
     \[
     \frac{1^2 + 20 \cdot 1 \cdot 2 + 2^2}{1^3 + 2^3} = \frac{1 + 40 + 4}{1 + 8} = \frac{45}{9} = 5
     \]
   - For \((m, n) = (5, 1)\):
     \[
     \frac{5^2 + 20 \cdot 5 \cdot 1 + 1^2}{5^3 + 1^3} = \frac{25 + 100 + 1}{125 + 1} = \frac{126}{126} = 1
     \]
   - For \((m, n) = (1, 5)\):
     \[
     \frac{1^2 + 20 \cdot 1 \cdot 5 + 5^2}{1^3 + 5^3} = \frac{1 + 100 + 25}{1 + 125} = \frac{126}{126} = 1
     \]

Conclusion:
The positive integers that can be written in the given form are \(1\) and \(5\).

The final answer is \(\boxed{1 \text{ and } 5}\)