Will you count the numbers?

It is easy to see, that the sum of the digits (Sd) can only be odd, if we have an odd number of odd digits.

In the case of the four digit numbers, this means, that we either have one or three odd digits and the rest of the digits are even.

We can choose both the odd {1,3,5,7,9} and the even digits {0,2,4,6,8} from the sets with 5 elements each, with repetition. We have to be careful with the 0 though, as it cannot be our first digit (as e.g. 0654 is not a four digit number).

1.) case: we have 3 odd digits and 1 even digit:

We can choose the 3 odd digits 5 × 5 × 5 = 125 ways and the non-zero even digits 4 ways, plus the 0 one way. Now, we can put the non-zero even digits to all four places, but the 0 can be one of the last three digits only.

Hence, the number of odd-Sd integers in this case:

$5^3 ×(4 × 4 + 1 × 3) = 125 × 19 = 2375$

2.) case: We have 1 odd and 3 even digits:

Once again, it makes sense to separate 2 possibilities (sub-cases):

a) In this case, the odd digit is our first digit, which gives us a convenient choice from 5 for each digit, meaning the number of relevant values is:

$5 × 5 × 5 × 5 = 625$

b) The odd digit is not the first digit of our number:

This means, that the first digit is even and cannot be 0, so we can choose it 4 ways, the other 3 digits 5 ways (each) and the place of the odd digit 3 ways:

$4 × 5 × 5 × 5 × 3 = 1500$ odd-Sd numbers here.

Therefore, our solution should be

$2375 + 625 + 1500 = \boxed {4500}$

$\begin{aligned} &\text{Let the number be } x_1 x_2 x_3 x_4\\ & 9\geq x_1 \geq 1 ;9\geq x_2, x_3, x_4, \geq 0, \text{where } x_1, x_2, x_3, x_4 \in N \\ &\text{Number of ways of choosing } x_1 = {9\choose 1}\\ &\text{Number of ways of choosing } x_2 = {10\choose 1}\\ &\text{Number of ways of choosing } x_3 = {10\choose 1}\\ &\text{CASE 1:If the sum of the 3 digits is odd, then } x_4 \text{ can be chosen in }{5\choose 1} \text{i.e. any odd digit to make sum even.}\\ &\text{CASE 2:If the sum of the 3 digits is even, then } x_4 \text{ can be chosen in }{5\choose 1} \text{i.e. any even digit to make sum even.}\\ &\text{Thus in any case there are 5 ways of choosing } x_4 \\ &\text{Thus total 4-digit numbers with sum of digits even is } = 9\times10\times10\times5 = \text{ 4500} \\ \end{aligned}$