Writing successively the natural numbers, we obtain the sequence

$$
12345678910111213141516171819202122 \ldots
$$

What is the digit that is in the $2009^{th}$ position of this sequence?
Observe that:

- the numbers from 1 to 9 occupy nine positions;
- the numbers from 10 to 99 occupy $2 \times 90=180$ positions;
- the numbers from 100 to 199 occupy $3 \times 100=300$ positions;
- the numbers from 200 to 299 occupy $3 \times 100=300$ positions;
- the numbers from 300 to 399 occupy $3 \times 100=300$ positions; etc.

$\underbrace{100, \ldots 199}_{3 \times 100=300}, \underbrace{200, \ldots, 299}_{3 \times 100=300}, \underbrace{300, \ldots, 399}_{3 \times 100=300}, \underbrace{400, \ldots, 499}_{3 \times 100=300}, \underbrace{500, \ldots, 599}_{3 \times 100=300}, \underbrace{600, \ldots, 699}_{3 \times 100=300}$

Thus, the digits used to write from 1 to 699 occupy $9+180+6 \times 300=1989$ positions, so we are missing $2009-1989=20$ positions. Since $20=3 \times 6+2$, we still need to write from 700 to 706, obtaining 21 positions, with the digit 6 occupying the 21st position. Therefore, it is the digit 0 that occupies the $2009^{th}$ position.