1048. Which is greater and by how much: the sum of all even two-digit numbers or the sum of all odd two-digit numbers?
$\triangle$ First, let's calculate the total number of two-digit numbers. There are 99 numbers in total from 1 to 99, including both single-digit and two-digit numbers. Among them, there are 9 single-digit numbers. Therefore, the number of two-digit numbers is $99-9=90$.

Out of the 90 two-digit numbers, 45 are even and 45 are odd. Let's divide all the two-digit numbers into pairs:

$$
10 \text { and } 11, \quad 12 \text { and } 13, \quad 14 \text { and } 15, \ldots, 98 \text { and } 99 .
$$

In each pair, the odd number is 1 greater than the even number. Since there are 45 such pairs, the sum of all odd two-digit numbers is 45 greater than the sum of all even two-digit numbers.

Answer: The second sum is greater by 45.