jmc

Setting $y=1,$ we get $$f(x)f(1)=f(x)+\frac{2005}{x}+2005^2.$$The value $f(1)$ cannot be 1, and so we can solve for $f(x)$ to get $$f(x)=\frac{2005/x+2005^2}{f(1)-1}.$$In particular, $$f(1)=\frac{2005+2005^2}{f(1)-1}.$$Then $f(1{)}^2-f(1)-2005^2-2005=0,$ which factors as $(f(1)-2006)(f(1)+2005)=0.$ Hence, $f(1)=2006$ or $f(1)=-2005.$

If $f(1)=2006,$ then $$f(x)=\frac{2005/x+2005^2}{2005}=\frac{1}{x}+2005.$$We can check that this function works.

If $f(1)=-2005,$ then $$f(x)=\frac{2005/x+2005^2}{-2006}.$$We can check that this function does not work.

Therefore, $$f(x)=\frac{1}{x}+2005,$$so $n=1$ and $s=\frac{1}{2}+2005=\frac{4011}{2},$ so $n\times s=\boxed{\frac{4011}{2}}.$