The polynomial $ x^3\minus{}2004x^2\plus{}mx\plus{}n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two. How many values of $ n$ are possible?

$ \textbf{(A)}\ 250,\!000 \qquad
\textbf{(B)}\ 250,\!250 \qquad
\textbf{(C)}\ 250,\!500 \qquad
\textbf{(D)}\ 250,\!750 \qquad
\textbf{(E)}\ 251,\!000$
1. Let the polynomial be \( P(x) = x^3 - 2004x^2 + mx + n \). Given that it has three distinct positive zeros, one of which is an integer and the sum of the other two, we can denote the roots as \( a, b, \) and \( c \) where \( c \) is the integer root and \( a + b = c \).

2. By Vieta's formulas, the sum of the roots \( a + b + c = 2004 \). Since \( c = a + b \), we substitute to get:
   \[
   a + b + (a + b) = 2004 \implies 2(a + b) = 2004 \implies a + b = 1002
   \]
   Therefore, \( c = 1002 \).

3. The polynomial can be factored as:
   \[
   P(x) = (x - a)(x - b)(x - 1002)
   \]

4. Expanding this product, we get:
   \[
   P(x) = x^3 - (a + b + 1002)x^2 + (ab + 1002a + 1002b)x - 1002ab
   \]
   Using \( a + b = 1002 \), we simplify:
   \[
   P(x) = x^3 - 2004x^2 + (ab + 1002a + 1002b)x - 1002ab
   \]

5. From Vieta's formulas, we know:
   \[
   m = ab + 1002a + 1002b \quad \text{and} \quad n = 1002ab
   \]

6. To find the number of possible values for \( n \), we need to determine the number of distinct integer pairs \((a, b)\) such that \( a + b = 1002 \) and \( a \neq b \).

7. Since \( a \) and \( b \) are distinct positive integers, we can write \( b = 1002 - a \). The product \( ab \) becomes:
   \[
   ab = a(1002 - a) = 1002a - a^2
   \]

8. The function \( f(a) = 1002a - a^2 \) is a quadratic function opening downwards, reaching its maximum at \( a = \frac{1002}{2} = 501 \). However, since \( a \) and \( b \) must be distinct, \( a \) cannot be 501.

9. The range of \( a \) is from 1 to 1001 (since \( a \neq b \)). For each \( a \) in this range, \( b = 1002 - a \) is also an integer. Therefore, there are 1001 possible pairs \((a, b)\).

10. Each pair \((a, b)\) gives a unique value of \( n = 1002ab \). Since \( a \) ranges from 1 to 1001, there are 1001 possible values for \( n \).

The final answer is \(\boxed{250,500}\).