jmc

Let $f(x)$ be the given expression. We first examine the possible values of $f(x)$ for $x$ in the interval $(0,1].$ Note that $f(0)=0,$ while $f(1)=2+4+6+8=20.$

As we increase $x$ from $0$ to $1,$ each of the four floor functions "jumps up" by $1$ at certain points. Furthermore, if multiple floor functions "jump up" at the same value of $x,$ then some integers will be skipped.

For each $k,$ the function $\lfloor kx\rfloor$ "jumps up" at $x=\frac{1}{k},\frac{2}{k},\dots ,\frac{k-1}{k},\frac{k}{k}.$ Therefore, we see that at $x=\frac{1}{2}$ and $x=1,$ all four of the given functions "jump up," so that three integers are skipped. Also, for $x=\frac{1}{4}$ and $x=\frac{3}{4},$ the functions $\lfloor 4x\rfloor$ and $\lfloor 8x\rfloor$ both "jump up," skipping one integer.

Thus, for $0<x\le 1,$ $f(x)$ takes $20-3-3-1-1=12$ positive integer values. Notice that $$\begin{array}{rl}f(x+1)& =\lfloor 2(x+1)\rfloor +\lfloor 4(x+1)\rfloor +\lfloor 6(x+1)\rfloor +\lfloor 8(x+1)\rfloor \\ & =\left(\lfloor 2x\rfloor +2\right)+\left(\lfloor 4x\rfloor +4\right)+\left(\lfloor 6x\rfloor +6\right)+\left(\lfloor 8x\rfloor +8\right)\\ & =f(x)+20.\end{array}$$Therefore, in the interval $1<x\le 2,$ $f(x)$ takes $12$ more integer values between $21$ and $40,$ respectively. In general, $f(x)$ takes $12$ out of every $20$ positive integer values from the list $20a,20a+1,\dots ,2a+19.$

Since $20$ is a divisor of $1000,$ exactly $\frac{12}{20}=\frac{3}{5}$ of the first $1000$ positive integers are possible values for $f(x).$ Thus the answer is $1000\cdot \frac{3}{5}=\boxed{600}.$