But it's floored!

This is a step function that makes 11 jumps on every interval $[n,n+1)$ , where $n$ is an integer, namely, at $n+\frac{k}{8}$ for $k=1,2,..,7$ and at $n+\frac{k}{6}$ , for $k=1,...,5$ .... but we need to subtract one since $\frac{4}{8}=\frac{3}{6}$ . Thus $f(x)$ attains 12 distinct values on $[n,n+1)$ .

Now $f(0)=0$ and $f(50)=100+200+300+400=1000$ . On the 50 intervals $[n,n+1)$ for $n=0,..,49$ , the function $f(x)$ attains $50\times12=600$ non-negative values $<1000$ . If we reject 0 and include 1000, we end up with $\boxed{600}$ positive values $\leq{1000}$ .