Yes it's bound to be even.

PROOF:

Every even natural number can be written as 2p, where p is a natural number.

Every odd natural number can be written as 2q-1, where q is a natural number.

Suppose M and N are natural numbers and the list contains

the odd number 2M-1 of even natural numbers 2p1,2p2,...,2p2M-1 

and

the even number 2N of odd natural numbers 2q1-1,2q2-1,...,2q2N-1.   

Then the sum of the numbers in the list = 

 +  =

 -  +  =

2 - 2M + 2

2

which is an even number since it has a factor of 2.

[Note: We used 2p-1 for odd numbers rather than the usual 2p+1 because since
p is a natural number, the smallest odd natural number which 2p+1 can be is
3 when p=1, so to include 1, and some of the numbers in the list could be 1,
we must use 2p-1. The final result is positive since the first summation in
the parentheses above is greater than M.]

Edwin