Determine all monotone functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$, so that for all $x, y \in \mathbb{Z}$
\[f(x^{2005} + y^{2005}) = (f(x))^{2005} + (f(y))^{2005}\]
1. **Initial Condition and Monotonicity**:
   We are given that \( f: \mathbb{Z} \rightarrow \mathbb{Z} \) is a monotone function satisfying the functional equation:
   \[
   f(x^{2005} + y^{2005}) = (f(x))^{2005} + (f(y))^{2005}
   \]
   Since \( f \) is monotone, it is either non-decreasing or non-increasing.

2. **Evaluating at Zero**:
   Let \( x = 0 \) and \( y = 0 \):
   \[
   f(0^{2005} + 0^{2005}) = (f(0))^{2005} + (f(0))^{2005}
   \]
   Simplifying, we get:
   \[
   f(0) = 2(f(0))^{2005}
   \]
   This implies:
   \[
   f(0) = 0 \quad \text{or} \quad f(0) = \frac{1}{2^{2004}}
   \]
   Since \( f \) maps integers to integers, \( f(0) = 0 \).

3. **Evaluating at Powers**:
   Let \( y = 0 \):
   \[
   f(x^{2005}) = (f(x))^{2005}
   \]
   This implies that \( f(x) \) must be such that raising it to the 2005th power gives the same result as applying \( f \) to \( x^{2005} \).

4. **Considering \( f(1) \)**:
   Let \( x = 1 \):
   \[
   f(1^{2005}) = (f(1))^{2005}
   \]
   Simplifying, we get:
   \[
   f(1) = (f(1))^{2005}
   \]
   The solutions to this equation are \( f(1) = 1 \) or \( f(1) = -1 \).

5. **Monotonicity and Further Values**:
   - If \( f(1) = 1 \):
     \[
     f(2) = f(1 + 1) = f(1^{2005} + 1^{2005}) = (f(1))^{2005} + (f(1))^{2005} = 1 + 1 = 2
     \]
     By induction, we can show that \( f(n) = n \) for all \( n \geq 0 \).

   - If \( f(1) = -1 \):
     \[
     f(2) = f(1 + 1) = f(1^{2005} + 1^{2005}) = (f(1))^{2005} + (f(1))^{2005} = (-1)^{2005} + (-1)^{2005} = -1 + -1 = -2
     \]
     By induction, we can show that \( f(n) = -n \) for all \( n \geq 0 \).

6. **Negative Values**:
   Since \( f \) is monotone and we have determined \( f \) for positive integers, we can extend this to negative integers by considering the property:
   \[
   f(-x) = -f(x)
   \]
   This ensures that \( f \) is an odd function.

7. **Conclusion**:
   The only monotone functions that satisfy the given functional equation are:
   \[
   f(x) = x \quad \text{and} \quad f(x) = -x
   \]

The final answer is \(\boxed{f(x) = x \text{ or } f(x) = -x}\)