107. Each of the integers from one to a billion inclusive is replaced by the sum of the digits of the number; single-digit numbers, of course, do not change, while the others decrease. Then each of the newly obtained numbers is again replaced by the sum of its digits, and so on until a billion single-digit numbers are obtained. Which numbers will be more among the resulting numbers - ones or twos?
107. It is a good fact to know that every positive integer $N$ gives the same remainder when divided by 9 as the sum of its digits. (This follows from the fact that the digit $a_{k}$ in the $(k+1)$-th position from the end in the decimal representation of $N$ represents the term $a_{k} \cdot 10^{k}$ in the expansion of $N$ in powers of ten; but $a_{k} \cdot 10^{k} = a_{k} \cdot (99 \ldots 9 + 1) = 99 \ldots 9 a_{h} + a_{h}$ gives the same remainder when divided by 9 as $a_{h}$.) From this, it follows that if the number $N$ gives a remainder $r \neq 0$ when divided by 9, then by repeatedly replacing the number with the sum of its digits, we will eventually arrive at a single-digit number $r$, and if $N$ is divisible by 9, we will arrive at the number 9. Thus, among our billion single-digit numbers, the ones will correspond to the numbers $1, 10, 19, 28, \ldots, 1000000000$, which give a remainder of 1 when divided by 9, and the twos will correspond to the numbers $2, 11, 20, 29, \ldots, 999999992$, which give a remainder of 2 when divided by 9. But the first set of numbers is obviously one more than the second set; therefore, we will get one more one than two.