Odd or even

Since there are 84 even numbers and 85 odd numbers then by pigeon hole principle, any one parenthesis contain two odd numbers, whatever the permutation may be whose difference yielding an even number. Any one even number in the product and the product is even.

Somewhat tricky solution to answer! First I need to point out a few facts.The option "None of these choices" cannot be true since the sum is either an even number or odd and both of which are expressed in the rest of the options.

Solution : Suppose -by false assumption- that the product CAN be an odd number,which means none of the parentheses should be even and all of them have to odd.Therefore we should have:

$\quad \quad \quad \quad \quad \quad a_1=2k \rightarrow a_1 - 1=2k-1 \text{(odd)}$

$\quad \quad \quad \quad \quad \quad a_2=2k+1 \rightarrow a_2-2=2k-1 \text{(odd)}$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad .$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad .$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad .$

$\quad \quad \quad \quad \quad \quad a_{169}=2k \rightarrow a_{169}-169=2k-169 \text{(odd)}$

$\text {Hence by counting all the even integers we see that there are a}$

$\text{total of 85 even integers needed for the product not to be even.}$

Now in the original sequence,however, we can easily see that the number of even integers are $84$ ,contradicting what we said in the line above.Thus,the proof is complete and the sum is always an even number.