29. All natural numbers from 1 to 100 inclusive are divided into 2 groups - even and odd. Determine in which of these groups the sum of all digits used to write the numbers is greater and by how much.
29. Odd numbers $1,3,5,7, \ldots, 99$; even numbers $2,4,6,8, \ldots, 100$. Let's compare the sums of the single-digit numbers in these groups: $1+3+5+7+9=25,2+$ $+4+6+8=20,25-20=5$. Consider the numbers with the tens digit 1. The sum of the digits will be greater in the group that has a greater sum of the units digits. As shown above, such a sum is greater in the group of odd numbers by 5. Similarly, for numbers with the tens digit $2,3, \ldots, 9$. Thus, the sum of all digits of odd numbers from 1 to 99 is greater than the sum of the digits of even numbers from 2 to 98 by $5 \cdot 10=50$. But the number 100 belongs to the group of even numbers, and the sum of its digits is 1, so $50-1=49$. Therefore, the sum of all digits used to write the odd numbers from 1 to 100 is 49 greater than the sum of the digits used to write the even numbers from 1 to 100.