Can a natural number have an odd number of the even divisors and an even number of the odd divisors?
Example: All the prime numbers have an even number of the odd divisors (2), but also 0 even divisors (we want odd number of even divisors and 0 is considered as even number)
9 has odd number (3) of odd divisors so it won't fit either
No, it cannot. The total number of divisors (and therefore the number of even divisors) must be a multiple of the number of odd divisors.
Say, for instance, that our number is divisible by 4 but not 8. Take all the odd divisors, and multiply them by $2$. You now have all the even divisors which are not divisible by $4$. Multiply them by $2$ again, and you have all the divisors which are divisible by $4$.
So we have here a partition of all the divisors into three equally-sized parts, one of which is exactly all the odd divisors. So the total number of divisors is three times the number of odd divisors, and the number of even divisors is twice that of odd divisors.
A corresponding argument works no matter how many times $2$ goes into our number.