$14 \cdot 46$ Among the first 1000 positive integers, how many can be expressed in the form $[2 x]+[4 x]+[6 x]+[8 x]$.
(3rd American Invitational Mathematics Examination, 1985)
[Solution 1] Let $f(x)=[2 x]+[4 x]+[6 x]+[8 x]$.
Suppose $n$ is an integer, then
$$
\begin{aligned}
& f(x+n) \\
= & {[2 x+2 n]+[4 x+4 n]+[6 x+6 n]+[8 x+8 n] } \\
= & {[2 x]+[4 x]+[6 x]+[8 x]+20 n } \\
= & f(x)+20 n .
\end{aligned}
$$

If some integer $k$ can be expressed as $f\left(x_{0}\right)$, that is
$$
f\left(x_{0}\right)=k .
$$

where $x_{0}$ is a certain real number.
Then by (1), for $n=1,2,3, \cdots$,
$$
k+20 n=f\left(x_{0}\right)+20 n=f\left(x_{0}+n\right) .
$$

That is, $k+20$ can also be expressed in the form of $f(x)$.
Therefore, we only need to pay attention to which of the first 20 positive integers can be produced by $f(x)$ when $x$ varies in $(0,1]$.

Furthermore, if $x$ increases, then $f(x)$ will only change when $2 x, 4 x, 6 x, 8 x$ reach some integer, and it will always turn into a larger integer.

In $(0,1]$, this change occurs precisely when $x$ has the form $\frac{m}{n}$, where $1 \leqslant m \leqslant n$, and $n=2,4,6,8$. There are 12 such fractions, and they are arranged from smallest to largest as follows:
$$
\frac{1}{8}, \frac{1}{6}, \frac{1}{4}, \frac{1}{3}, \frac{3}{8}, \frac{1}{2}, \frac{5}{8}, \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, 1 .
$$

Therefore, only 12 of the first 20 positive integers can be expressed in the required form.
Since $1000=50 \cdot 20$, by (1), the first 1000 positive integers are divided into 50 segments, each consisting of 20 consecutive positive integers, of which 12 can be expressed in the required form.
Thus, the number of integers that meet the requirement is $50 \cdot 12=600$.
[Solution 2] Let $f(x)=[2 x]+[4 x]+[6 x]+[8 x]$.
From this, we can derive
$$
\begin{aligned}
f\left(x+\frac{1}{2}\right) & =[2 x+1]+[4 x+2]+[6 x+3]+[8 x+4] \\
& =[2 x]+[4 x]+[6 x]+[8 x]+10 \\
& =f(x)+10 .
\end{aligned}
$$

We only need to pay attention to which natural numbers $f(x)$ can express when $x \in\left[0, \frac{1}{2}\right)$.
When $x \in\left[0, \frac{1}{2}\right)$, $[2 x]=0$, thus
$$
f(x)=[4 x]+[6 x]+[8 x] .
$$

We discuss the following cases:
(1) When $x \in\left[0, \frac{1}{8}\right)$, at this time
$$
f(x)=0 .
$$
(2) When $x \in\left[\frac{1}{8}, \frac{1}{6}\right)$, at this time $[4 x]=[6 x]=0$,
$$
f(x)=[8 x]=1,
$$
(3) When $x \in\left[\frac{1}{6}, \frac{1}{4}\right)$, at this time $[4 x]=0$,
$$
f(x)=[6 x]+[8 x]=1+1=2,
$$
(4) When $x \in\left[\frac{1}{4}, \frac{1}{3}\right)$, at this time
$$
f(x)=1+1+2=4,
$$
(5) When $x \in\left[\frac{1}{3}, \frac{3}{8}\right)$, at this time
$$
f(x)=1+2+2=5,
$$
(6) When $x \in\left[\frac{3}{8}, \frac{1}{2}\right)$, at this time
$$
f(x)=1+2+3=6 \text {, }
$$
(7) When $x=\frac{1}{2}$, at this time
$$
f(x)=10 .
$$

Therefore, when $x \in\left(0, \frac{1}{2}\right]$, $f(x)$ can express the natural numbers $1,2,4,5,6,10$.
These numbers plus multiples of 10 can also be expressed in this form, so among the first 1000 natural numbers, $6 \cdot 100=600$ can be expressed in this form.