jmc

Note that $x\lfloor x\rfloor$ is strictly increasing in $x.$ Let $a$ be an integer. Given that $\lfloor x\rfloor =a,$ we have $a\le x<a+1,$ so $$a^2\le x\lfloor x\rfloor <a^2+a.$$Therefore, $x\lfloor x\rfloor$ takes $$(a^2+a-1)-a^2+1=a$$integer values over all $x$ such that $\lfloor x\rfloor =a.$

Note that $x\lfloor x\rfloor =1$ when $x=1,$ and that if $x\lfloor x\rfloor <1000,$ then $\lfloor x{\rfloor }^2<1000,$ so $a\le 31.$ For $a=1,2,\dots ,31,$ we get $$1+2+\cdots +31=\frac{31\cdot 32}{2}=496$$integer values of $x\lfloor x\rfloor .$ For $a\ge 32,$ we have $x\lfloor x\rfloor \ge a^2=1024>1000,$ so we do not get any more values of $n.$ The answer is $\boxed{496}.$