Solution below.

Reveal solution by highlighting:

We have to count some of the integers from 100 to 1000, not included. There are 1000-100-1 = 899 such integers in all. But we only want those, where the sum of the digits is even.

For a moment, let’s include 100 in our integers. And let’s look at some random ab0-ab9. (Remember a cannot be 0.) We can group these 10 integers as ab0-ab8 and ab1-ab9, depending on whether the last digit is even or odd. There are 5 integers in each group. Either one group or the other must have the same characteristic for all 5 integers: the sum of the digits is even for all 5. (For the other group, the sum is always odd.) That’s 900 integers done, and half of them have an even digit sum, that’s 450.

But remember, we included 100. But that’s okay, 100 has an odd digit sum, so we can just throw it away now. So the answer is 450.