Question:
The polynomial $x^3 - 2004 x^2 + mx + n$ 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?

Answer:
Let $a$ denote the zero that is an integer.  Because the coefficient of $x^3$ is 1, there can be no other rational zeros, so the two other zeros must be $\frac{a}{2} \pm r$ for some irrational number $r$.  The polynomial is then \[(x-a) \left( x - \frac{a}{2} - r \right) \left( x - \frac{a}{2} + r \right) = x^3 - 2ax^2 + \left( \frac{5}{4}a^2 - r^2 \right) x - a \left( \frac{1}{4}a^2 - r^2 \right).\]Therefore $a=1002$ and the polynomial is \[x^3 - 2004 x^2 + (5(501)^2 - r^2)x - 1002((501)^2-r^2).\]All coefficients are integers if and only if $r^2$ is an integer, and the zeros are positive and distinct if and only if $1 \leq r^2
\leq 501^2 - 1 = 251000$.  Because $r$ cannot be an integer, there are $251000 - 500 = \boxed{250500}$ possible values of $n$.