### Instruction: How many of the first $1000$ positive integers can be expressed in the form
\[\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor\]where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$?### Response: I notice that the expression involves only even integers, so it must be even for any $x$.
That means I can ignore all the odd numbers in the first $1000$ positive integers, since they cannot be expressed in this form.
So I only need to consider the even numbers, of which there are $500$.
Now I wonder how the expression changes as $x$ increases.
I see that each term in the expression is a linear function of $x$ with a slope of $2, 4, 6, 8$, respectively.
So the whole expression has a slope of $20$.
That means that as $x$ increases by $0.05$, the expression increases by $1$.
So if I start from $x = 0$, where the expression is $0$, and increase $x$ by $0.05$ repeatedly, I will get all the even integers from $0$ to $1000$.
For example, when $x = 0.1$, the expression is $2 + 4 + 6 + 8 = 20$.
When $x = 0.2$, the expression is $4 + 8 + 12 + 16 = 40$.
And so on.
So I conclude that every even integer from $0$ to $1000$ can be expressed in this form, as long as $x$ is a multiple of $0.05$.
Therefore, the answer is $500$.