16. 1.1 $\star$ Calculate the sum of digits for each number from 1 to $10^{9}$, resulting in $10^{9}$ new numbers, then find the sum of digits for each of these new numbers; continue this process until all numbers are single digits. In the final set of numbers, are there more 1s or 2s?
When a positive integer and the sum of its digits are divided by 9, the remainder is the same. Therefore, when the final single digit is 1, the original number, when divided by 9, leaves a remainder of 1, i.e., the original number is $1, 10, 19, \cdots, 999999991, 10^{9}$.

Similarly, when the final single digit is 2, the original number is $2, 11, 20, \cdots, 999999992$. Clearly, there is one more number that leaves a remainder of 1. Therefore, among the final single digits, 1 is more frequent.