For two positive integers a and b, which are relatively prime, find all integer that can be the great common divisor of $a+b$ and $\frac{a^{2005}+b^{2005}}{a+b}$.
1. **Given**: Two positive integers \(a\) and \(b\) are relatively prime, i.e., \(\gcd(a, b) = 1\). We need to find all integers that can be the greatest common divisor (GCD) of \(a+b\) and \(\frac{a^{2005} + b^{2005}}{a+b}\).

2. **Expression Simplification**: Let \(d = \gcd(a+b, \frac{a^{2005} + b^{2005}}{a+b})\). We need to find \(d\).

3. **Euclidean Algorithm**: We use the Euclidean algorithm to find the GCD. First, we note that:
   \[
   \gcd(a+b, \frac{a^{2005} + b^{2005}}{a+b}) = \gcd(a+b, a^{2005} + b^{2005})
   \]
   This is because \(\frac{a^{2005} + b^{2005}}{a+b}\) is an integer.

4. **Polynomial Division**: We can express \(a^{2005} + b^{2005}\) in terms of \(a+b\):
   \[
   a^{2005} + b^{2005} = (a+b)(a^{2004} - a^{2003}b + a^{2002}b^2 - \cdots + b^{2004})
   \]
   Therefore,
   \[
   \frac{a^{2005} + b^{2005}}{a+b} = a^{2004} - a^{2003}b + a^{2002}b^2 - \cdots + b^{2004}
   \]

5. **GCD Calculation**: We need to find \(\gcd(a+b, a^{2004} - a^{2003}b + a^{2002}b^2 - \cdots + b^{2004})\). Using the Euclidean algorithm:
   \[
   \gcd(a+b, a^{2004} - a^{2003}b + a^{2002}b^2 - \cdots + b^{2004}) = \gcd(a+b, -2a^{2003}b + \cdots + b^{2004})
   \]
   Continuing this process, we eventually get:
   \[
   \gcd(a+b, 2005b^{2004})
   \]

6. **Prime Factorization**: Since \(a\) and \(b\) are relatively prime, \(\gcd(a+b, b^{2004}) = 1\). Therefore:
   \[
   \gcd(a+b, 2005b^{2004}) = \gcd(a+b, 2005)
   \]

7. **Prime Factors of 2005**: The prime factorization of 2005 is:
   \[
   2005 = 5 \times 401
   \]
   Hence, the possible values for \(\gcd(a+b, 2005)\) are the divisors of 2005, which are:
   \[
   1, 5, 401, 2005
   \]

8. **Verification**: We can verify that these values are achievable by choosing appropriate values of \(a\) and \(b\). For example:
   - If \(a+b = 1\), then \(d = 1\).
   - If \(a+b = 5\), then \(d = 5\).
   - If \(a+b = 401\), then \(d = 401\).
   - If \(a+b = 2005\), then \(d = 2005\).

The final answer is \(\boxed{1, 5, 401, 2005}\)