How many of the first 1000 positive integers can be expressed as:
$$
\lfloor 2 x\rfloor+\lfloor 4 x\rfloor+\lfloor 6 x\rfloor+\lfloor 8 x\rfloor \text { of }
$$

form?
where $\mathbf{x}$ is some real number, and $\mathrm{Lz}\rfloor$ denotes the greatest integer not exceeding $z$.
$$
\begin{array}{l}
\text { Let } f(x)=\lfloor 2 x\rfloor+\lfloor 4 x\rfloor \\
+\lfloor\lfloor 6 x\rfloor+\lfloor 8 x\rfloor .
\end{array}
$$

It can be seen that if $n$ is an integer, then from (1) we have:
$$
f(x+n)=f(x)+20 n .
$$

If some integer $k$ can be expressed as $f\left(x_{0}\right)$, where $x_{0}$ is some real number, then by (2), for $n=1,2,3, \cdots$, $k+20 n=f\left(x_{0}\right)+20 n=f\left(x_{0}+n\right)$, i.e., $k+20 n$ can also be expressed in the form (i). Therefore, we focus on the values that $f(x)$ can produce when $x$ varies in $(0,1]$. Furthermore, if $x$ increases, $f(x)$ will only change when $2 x, 4 x, 6 x, 8 x$ reach some integer value, and it will always change to a larger integer. In $(0,1]$, this change occurs precisely when $x$ has the form $m / n$, where $1 \leqslant m \leqslant n$, and $n=2,4,6$ or 8. There are 12 such fractions, arranged from smallest to largest:
$$
\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3},
$$
$\frac{3}{4}, \frac{5}{6}, \frac{7}{8}$ and 1.
Thus, only 12 of the first 20 positive integers can be expressed in the required form, because $1000=50 \cdot 20$, by (2) we know that the first 1000 positive integers are divided into 50 segments, each consisting of 20 consecutive positive integers, of which only 12 can be expressed in the required form. Therefore, the total number of integers that meet the requirement is $50 \times 12=600$.