$9 \cdot 8$ For all positive integers from 1 to 1000000000, find the sum of the digits of each number, then for the resulting 1 billion numbers, find the sum of the digits of each number, $\cdots$, until obtaining 1 billion single-digit numbers. Ask whether there are more 1s or 2s in the resulting numbers?
(4th All-Russian Mathematical Olympiad, 1964)
[Solution] Since any positive integer and the sum of its digits have the same remainder when divided by 9.
Therefore, the numbers whose digit sum is 1 are all numbers of the form $9k+1$, i.e.,
$$
1,10,19,28, \cdots, 999999991,1000000000 .
$$

There are 111111112 in total.
And the numbers whose digit sum is 2 are all numbers of the form $9k+2$, i.e.,
$$
2,11,20,29, \cdots, 999999992 \text {. }
$$

There are 111111111 in total.
Therefore, the numbers with a digit sum of 1 are 1 more than those with a digit sum of 2.