jmc

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 $$(x-a)\left(x-\frac{a}{2}-r\right)\left(x-\frac{a}{2}+r\right)=x^3-2a{x}^2+\left(\frac{5}{4}{a}^2-r^2\right)x-a\left(\frac{1}{4}{a}^2-r^2\right).$$Therefore $a=1002$ and the polynomial is $$x^3-2004x^2+(5(501{)}^2-r^2)x-1002((501{)}^2-r^2).$$All coefficients are integers if and only if $r^2$ is an integer, and the zeros are positive and distinct if and only if $1\le r^2\le 501^2-1=251000$. Because $r$ cannot be an integer, there are $251000-500=\boxed{250500}$ possible values of $n$.