For a natural number $n$, let the sum of its digits be denoted as $a_{n}$, for example,
$$
\begin{array}{l}
a_{2009}=2+0+0+9=11, \\
a_{2010}=2+0+1+0=3 .
\end{array}
$$

Then $a_{1}+a_{2}+\cdots+a_{2010}=$ ( ).
(A) 28062
(B) 28065
(C) 28067
(D) 28068
6. D.

Consider all natural numbers from 1 to 2010 as four-digit numbers (if $n$ is less than four digits, add 0s at the beginning to make it four digits, which does not change the value of $a_{n}$).

Notice that, the number of times 1 appears in the thousands, hundreds, tens, and units place are $10^{3}$, $2 \times 10^{2}$, $2 \times 10^{2}+1$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 1 appears is
$$
10^{3}+2 \times 10^{2} \times 3+2=1602 \text {. }
$$
The number of times 2 appears in the thousands, hundreds, tens, and units place are $11$, $2 \times 10^{2}$, $2 \times 10^{2}$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 2 appears is
$$
11+2 \times 10^{2} \times 3+1=612 \text {. }
$$

Similarly, the total number of times $k(k=3,4, \cdots, 9)$ appears is $2 \times 10^{2} \times 3+1=601$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{2010} \\
=1602 \times 1+612 \times 2+601(3+4+\cdots+9) \\
=28068 \text {. }
\end{array}
$$