All-Soviet-Union Mathematical Olympiad

Taking digit sums repeatedly gives the remainder after dividing the number by $9$, or $9$ if the number is exactly divisible by $9$. $10^9 - 1 = 9n$, and for any $r \geq 0$ the nine consecutive numbers $9r + 1$, $9r + 2$, ..., $9r + 9$ include just one number giving remainder $1$ and one number giving remainder $2$. Hence the numbers up to $10^9 - 1$ give equal numbers of $1$s and $2$s. $10^9$ itself gives $1$, so there is just one more of the $1$s than the $2$s.