17.1.9 ** $x$ is a real number, how many of the first 1000 positive integers can be expressed in the form $[2 x]+[4 x]+[6 x]+[8 x]$?
Let $f(x)=[2 x]+[4 x]+[6 x]+[8 x]$.
When $x$ varies in $(0,1]$, the value of $f(x)$ changes at 12 "critical points": $\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3^{\prime}}$, $\frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}$, and 1, so $f(x)$ takes 12 different positive integer values in 1-20.

On the other hand, $f(x+n)=f(x)+20 n$ holds for any positive integer $n$. Dividing $1-1000$ into 50 segments, each of 20 numbers. In each segment, $f(x)$ can take 12 values. Therefore, a total of $50 \times 12=600$ values can be taken, meaning that 600 of the first 1000 positive integers can be expressed in the form $[2 x]+[4 x]+[6 x]+[8 x]$.