With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables.  However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.

Therefore, our best approach is to simply analyze each answer choice.

$\displaystyle m^2n$:  Since $\displaystyle m$ is odd, $\displaystyle m^2$ is also odd, since and odd number multiplied by an odd number yields an odd product.  Since $\displaystyle n$ is also odd, multiplying it by $\displaystyle m^2$ will again yield an odd product, so this expression is always odd.

$\displaystyle m-2n$:  Since $\displaystyle n$ is odd, multiplying it by 2 will yield an even number.  Subtracting this number from $\displaystyle m$ will also give an odd result, since an odd number minus an even number gives an odd number.  Therefore, this answer is also always odd.

$\displaystyle mn$:  Since both numbers are odd, their product will also always be odd.

$\displaystyle 2m-n$:  Since $\displaystyle m$ is odd, multiplying it by 2 will give an even number.  Since $\displaystyle n$ is odd, subtracting it from our even number will give an odd number, since an even number minus and odd number is always odd.  Therefore, this answer will always be odd.

$\displaystyle \frac{m+n}{2}$:   Since both numbers are odd, there sum will be even.  However, dividing an even number by another even number (2 in our case) does not always produce an even or an odd number.  For example, 5 and 7 are both odd.  Their sum, 12, is even.  Dividing by 2 gives 6, an even number.  However, 5 and 9 are also both odd.  Their sum, 14, is even, but dividing by 2 gives 7, an odd number.  Therefore, this expression isn't necessarily always odd or always even, and is therefore our answer.

To find the answer to this question, calculate the total jelly beans for each person:

Portia: $\displaystyle 3$ * <Theodore's count of jelly beans>, which is $\displaystyle 3 * 15$ or $\displaystyle 45$

Harvey: $\displaystyle 5$ * <Portia's count of jelly beans>, which is $\displaystyle 5*45$ or $\displaystyle 225$

So, the total is:

$\displaystyle 15+45+225=285$

(Do not forget that you need those original $\displaystyle 15$ for Theodore!)

Rewrite the product into the expression $\displaystyle 13\cdot 7$, and multiply the ones digit of both numbers.

$\displaystyle 3(7)=21$

The ones digit of the final answer is the ones digit from the multiplication of 3 and 7 thus the 1.  The tens digit, 2, will be carried over to the next calculation.

Multiply the tens digit of 13 to 7 and add the carry over.

$\displaystyle 7(1)+2=9$ 

The value of 9 is the tens digit of the final answer.  

Combine the tens digit with the ones digit.  The answer is $\displaystyle 91$.

For the product of two numbers to be even, one number must be even. For the product of two numbers to be odd, both numbers must be odd.

Remember:

$\displaystyle O \cdot E = E$

$\displaystyle O \cdot O = O$

$\displaystyle E \cdot E = E$