Short and Sweet

All possible 5 digit numbers would be of the form abcde where a $\in\{1,2,3...,9\}$ and b,c,d,e $\in\{0,1,2,3...,9\}$ .

Observe that if the sum a+b+c+d is even, then e (the last digit) can take the values 0,2,4,6 and 8 and if the sum is odd, then e can take the values 1,3,5,7 and 9 such that sum a+b+c+d+e is even.

Therefore, e takes exactly 5 values depending on whether the sum a+b+c+d is odd or even.

Therefore, required number of 5 digit numbers= $9\times10\times10\times10\times5=45000.$

For the sum of digits of a 5-digit number be even, there are only three cases: one, three or five of the digits are even. But the first (left-most) digit cannot be 0. Since there are five odd digits and five even digits. Every digit of the 5-digit number has five combinations except for the first digit has only four combinations when it is even. Therefore, we have:

• With one digit even, $N_1 = {1 \choose 1} 4 \times 5^4 + {4 \choose 1} 5^5 = 2500 + 12500 = 15000$
• With three digits even, $N_3 = {4 \choose 2} 4 \times 5^4 + {4 \choose 3} 5^5 = 15000+12500 = 27500$
• With five digits even, $N_5 = 4 \times 5^4 = 2500$

Therefore, the total number of 5-digit numbers whose sum of digits is even $N=N_1+N_3+N_5 = \boxed{45000}$