How many different ways are there to write 2004 as a sum of one or more positive integers which are all "aproximately equal" to each other? Two numbers are called aproximately equal if their difference is at most 1. The order of terms does not matter: two ways which only differ in the order of terms are not considered different.
1. Let's denote the number of terms as \( K \). We need to write 2004 as a sum of \( K \) terms that are approximately equal. This means the terms can either be \( n \) or \( n+1 \) for some integer \( n \).

2. Suppose there are \( a \) terms equal to \( n+1 \) and \( K-a \) terms equal to \( n \). The sum of these terms is:
   \[
   a(n+1) + (K-a)n = 2004
   \]

3. Simplify the equation:
   \[
   an + a + Kn - an = 2004
   \]
   \[
   Kn + a = 2004
   \]

4. From the equation \( Kn + a = 2004 \), we can solve for \( a \):
   \[
   a = 2004 - Kn
   \]

5. Since \( a \) must be a non-negative integer and \( 0 \le a < K \), we have:
   \[
   0 \le 2004 - Kn < K
   \]

6. This inequality can be split into two parts:
   \[
   0 \le 2004 - Kn \quad \text{and} \quad 2004 - Kn < K
   \]

7. The first part \( 0 \le 2004 - Kn \) implies:
   \[
   Kn \le 2004
   \]
   \[
   n \le \frac{2004}{K}
   \]

8. The second part \( 2004 - Kn < K \) implies:
   \[
   2004 < Kn + K
   \]
   \[
   2004 < K(n + 1)
   \]
   \[
   \frac{2004}{K} < n + 1
   \]
   \[
   \frac{2004}{K} - 1 < n
   \]

9. Combining the two inequalities, we get:
   \[
   \frac{2004}{K} - 1 < n \le \frac{2004}{K}
   \]

10. Since \( n \) is an integer, there is exactly one integer \( n \) that satisfies this inequality for each \( K \). Therefore, for each \( K \) from 1 to 2004, there is exactly one solution.

11. Since \( K \) can range from 1 to 2004, there are 2004 different ways to write 2004 as a sum of one or more positive integers that are approximately equal to each other.

The final answer is \(\boxed{2004}\).