Let me first thank the author for providing the sudokus.
I'm quite new to these competitions, that might also be why I missed something, as I cannot see why there is a correct solution to one of the simpler sudokus: all odd/even
Let me explain, and hopefully one of you can prove me wrong :)

The grey fields either contain all even or all odd numbers. There are 5 odd numbers (1,3,5,7,9) and 4 even numbers (2,4,6,8) possible. Look at the rightmost column, there are 3 empty grey fields and even numbers 2 and 6 already present. That means that the 3 grey fields in this column cannot be filled with all even numbers (as there are only two left: 4 and 8). Hence, all grey fields should be odd-numbered.
Now look at the second row, there are only two non-grey empty fields, which are located in the top-middle 3x3 square. Number 4 is still missing in this row, hence it should be filled in somewhere in the middle 3, however there is already a 4 present in the top-middle 3x3 square. Hence, the grey fields cannot all be odd-numbered. A contradiction!

edit:
oops... I now see that the restriction on odd/even should be interpreted per 3x3 square... sorry

Edited by gjdv 2012-07-12 2:01 AM