Question 543272
<pre>
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 2p<sub>1</sub>,2p<sub>2</sub>,...,2p<sub>2M-1</sub> 

and

the even number 2N of odd natural numbers 2q<sub>1</sub>-1,2q<sub>2</sub>-1,...,2q<sub>2N</sub>-1.   

Then the sum of the numbers in the list = 

{{{sum((2p[i]-1),i=1,2M)}}} + {{{sum((2q[j]),j=1,2N-1)}}} =

{{{sum((2p[i]),i=1,2M)}}} - {{{sum((1),i=1,2M)}}} + {{{sum((2q[j]),j=1,2N-1)}}} =

2{{{sum((p[i]),i=1,2M)}}} - 2M + 2{{{sum((q[j]),j=1,2N-1)}}}

2{{{(sum((p[i]),i=1,2M) - M + sum((q[j]),j=1,2N-1))}}}

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</pre>