$10 \cdot 46$ Find the sum of all digits of the numbers $1,2,3, \cdots, 10^{n}-2,10^{n}-1$. (Kiev Mathematical Olympiad, 1970)
[Solution] Correspond the number $m$ with $10^{n}-1-m$. Since
$$
m+\left(10^{n}-1-m\right)=10^{n}-1=\underset{n \uparrow}{99 \cdots 9},
$$

the sum of the digits of each pair of numbers $m, 10^{n}-1-m$ is $9n$.
Taking $m=1,2, \cdots, 10^{n}-2$, there are $\frac{10^{n}-2}{2}$ pairs of numbers whose sum of digits is $9n$.
Also, when $m=10^{n}-1$, there is only one number, and the sum of its digits is $9n$.
Thus, the sum of all digits of $1,2, \cdots, 10^{n}-2,10^{n}-1$ is
$$
9 n \cdot \frac{10^{n}-2}{2}+9 n=\frac{1}{2} \cdot 10^{n} \cdot 9 n \text {. }
$$