The problem has been solved already, so I will solve a different problem.
Note first that in the original problem the $1000$ kind of sticks out. So I would have preferred dealing with $1$ to $999$. (We can always deal with $1000$ at the end. And to tell the truth I really prefer $0$ to $999$.)
Let's bump the problem up a bit, by going from $0$ to $9999$.
For the sake of symmetry, "pad" the little numbers with initial $0$'s, so that everybody has $4$ digits.
The reason I am doing this is to make stuff as symmetrical as possible: symmetry is our friend.
So the numbers are
and so on, ending with
Think of any digit, like $7$, or even $0$.
How many $7$'s in total in the units place? Once we have put a $7$ there, we can fill out the rest of the spots in $7^3$ ways.
How many $7$'s in the next place? Again, once you have placed the $7$, there are $10^3$ ways to fill out the rest. Continue. We find that the total number of $7$'s is
$$4 \times 10^3.$$
This is also the total number of $0$'s! But some of the zeros have been obtained from the "padding" and should now be removed.
The numbers $0$ to $9$ have a padding of $0$ in the tens place, for a total of $10^1$. The ones from $0$ to $99$ have padding in the hundreds place, for a total of $10^2$. And the numbers $0$ to $999$ have padding in the thousands place, a total of $10^3$. So the number of genuine legitimate $0$'s is
$$4 \times 10^3 -(10^1+10^2+10^3).$$
In the case of your problem, the corresponding expression would be
$3\times 10^2 -(10^1+10^2)$. This is $190$. But your list started at $1$, which eliminates one $0$, and ended at $1000$, which adds $3$, for a total of $192$.
Generalization: It is easy now to see what the counts would be if we listed, say, all numbers from $0$ to $10^k-1$.