$1 \cdot 10$ Among the first 1000 positive integers, how many can be expressed in the form $[2 x]+[4 x]$ $+[6 x]+[8 x]$? where $x$ is some real number.
[Solution 1] Let
$$
f(x)=[2 x]+[4 x]+[6 x]+[8 x],
$$

then we have $f\left(x+\frac{1}{2}\right)=f(x)+10$. Therefore, if a positive integer can be expressed by the value of $f(x)$, then the integer plus 10 can also be expressed by the value of $f(x)$. Thus, it is sufficient to consider the first 10 natural numbers that can be expressed by $f(x)$.
Note that $f\left(\frac{1}{2}\right)=10$. When $x \in\left(0, \frac{1}{2}\right)$, $[2 x]=0$, so we have
$$
f(x)=[4 x]+[6 x]+[8 x].
$$

It is easy to see that
(1) When $x \in\left(0, \frac{1}{8}\right)$, $f(x)=0$;
(2) When $x \in\left[\frac{1}{8}, \frac{1}{6}\right)$, $f(x)=1$;
(3) When $x \in\left[\frac{1}{6}, \frac{1}{4}\right)$, $f(x)=2$;
(4) When $x \in\left[\frac{1}{4}, \frac{1}{3}\right)$, $f(x)=4$;
(5) When $x \in\left[\frac{1}{3}, \frac{3}{8}\right)$, $f(x)=5$;
(6) When $x \in\left[\frac{3}{8}, \frac{1}{2}\right)$, $f(x)=6$.

It can be seen that when $x \in\left(0, \frac{1}{2}\right]$, $f(x)$ can express the 6 natural numbers $1,2,4,5,6,10$. Therefore, among the first 1000 natural numbers, $f(x)$ can express 600 of them.
[Solution 2] Let
$$
\begin{array}{c}
f_{1}(x)=[2 x], f_{2}(x)=[4 x], f_{3}(x)=[6 x], f_{4}(x)=[8 x], \\
f(x)=f_{1}(x)+f_{2}(x)+f_{3}(x)+f_{4}(x).
\end{array}
$$

When the value of $f_{j}(x)$ increases by 1 as $x$ increases from small to large and reaches a certain point, we call that point an increment point, and denote the set of all increment points of $f_{j}(x)$ in $(0,50]$ as $M_{j}, j=1,2,3,4$. It is easy to see that
$$
\begin{array}{l}
M_{1}=\left\{\left.\frac{i}{2} \right\rvert\, i=1,2, \cdots, 100\right\}, \\
M_{2}=\left\{\left.\frac{i}{4} \right\rvert\, i=1,2, \cdots, 200\right\}, \\
M_{3}=\left\{\left.\frac{i}{6} \right\rvert\, i=1,2, \cdots, 300\right\}, \\
M_{4}=\left\{\left.\frac{i}{8} \right\rvert\, i=1,2, \cdots, 400\right\}.
\end{array}
$$

It can be seen that $M_{1} \subset M_{2} \subset M_{4}, M_{1} \subset M_{3}, M_{2} \cap M_{3}=M_{1}=M_{3} \cap M_{4}$.
Clearly, when and only when a point is an increment point of two $f_{j}(x)$, the value of $f(x)$ will increase by 2, i.e., it will skip 1 natural number that cannot be expressed. Similarly, when a certain $x$ is an increment point of 3 or 4 $f_{j}(x)$, the value of $f(x)$ will skip 2 or 3 natural numbers. From the above analysis, we know that each point in $M_{1}$ is an increment point of 4 $f_{j}(x)$, and they skip a total of 300 natural numbers. In addition, the points in $M_{2}-M_{1}$ are increment points of both $f_{2}(x)$ and $f_{4}(x)$, and they skip a total of 100 natural numbers. It is easy to see that $f(x)$ has no other increment points. Therefore, $f(x)$ can express 600 natural numbers.