I'm invoking the Fractional-Part Function!

since we are given q is non integer, we easily deduce $\dfrac{n!}{2000}$ is non integer. but for n≥ 15 it will always be an int. hence,we have to check the 1 ->14. what is fractional part, it is just $\dfrac{n!\pmod{2000}}{2000}$ we search for mod by multiplying to the las number $\begin{array}{c}11!&&\equiv&&1&&\equiv&& 1&&\pmod{2000}\\ 2!&&\equiv&& 1*2&&\equiv&& 2 &&\pmod{2000}\\ 3!&&\equiv&& 2*3&&\equiv &&6 &&\pmod{2000}\\ 4!&&\equiv&& 6*4&&\equiv&& 24&& \pmod{2000}\\ 5!&&\equiv&& 24*5&&\equiv&& 120&& \pmod{2000}\\ 6!&&\equiv&& 120*6 &&\equiv&& 720&& \pmod{2000}\\ 7!&&\equiv&& 720*7&&\equiv&& 40 &&\pmod{2000}\\ 8!&&\equiv&& 40*8&&\equiv&& 320 &&\pmod{2000}\\ 9!&&\equiv&& 320*9 &&\equiv&& 880 &&\pmod{2000}\\ 10!&&\equiv&& 880*10&&\equiv&& 800&& \pmod{2000}\\ 11!&&\equiv&& 800*11&&\equiv&&800 &&\pmod{2000}\\ 12!&&\equiv&& 800*12&&\equiv&& 1600 &&\pmod{2000}\\ 13!&&\equiv&& 1600*13&&\equiv&& 800&& \pmod{2000}\\ 14!&&\equiv&& 800*14&&\equiv&& 1200&& \pmod{2000}\\ \end{array}$ we see that q must be rational ,so must its fractional part for which $\sqrt{\dfrac{n!\pmod{2000}}{2000}}$ must be rational. we see that only 6 and 8 satisfies this each producing 2000 q, hence we have $2*2000=\boxed{4000}$ solutions. namely $(6,0.6),(6,1.6)....(6,1999.6)\quad and\quad(8,0.4),(8,1.4).....(8,1999.4)$