This problem requires only one simple calculation; namely, that half the balls in the box are odd and half even.
One principle of random selection from a box is that only the proportion matters: the behavior (of sampling with replacement) of a box with 100 balls is the same as that with just two balls, one odd and one even, because in both cases exactly half the balls are odd and half even.
This problem enjoys a symmetry: if we were to relabel all the even balls as odd and all the odd balls as even, the proportions of each in the box would not change but the parity of the sum would change. Consequently, the chance of an odd sum must equal the chance of an even sum. Since that's all that can occur, each chance must be $1/2$.
Different reasoning is needed when drawing an even number of balls, for then the parity of the sum does not change after the relabeling. For example, if originally three odd balls and one even ball had been drawn (for an odd sum), then after relabeling, three even balls and one odd ball are drawn--again for an odd sum. One way to draw a conclusion in this case is to consider the situation just before the last ball is drawn. At this point an odd number of balls have come out, whence there is a $1/2$ chance that their sum is odd, $1/2$ that it is even. The final ball will either switch the parity (if it's odd) or not (if it's even), but whether it switches or not doesn't create any imbalance in the chances.
This is kind of interesting, because pursuing this argument (inductively) to its end leads to a curious situation. As before, suppose the box starts out with equal numbers of odd and even balls. However, after each step (when you draw one ball at random, observe its parity, and replace it), someone else adds to or removes from the box an arbitrary number of balls unknown to you. (All you know is that at least one ball remains in the box.) This changes the chances to completely unknown values, anywhere between $0$ and $1$. Let's further suppose that a third party, who has not observed these proceedings, gets to decide when you stop drawing, so you don't even know in advance how many balls ($n$) you will draw, except that it is nonzero. The parities of the final sum still have equal chances of $1/2$. The reason should now be obvious: the parity of the sum of the last $n-1$ balls could be anything; adding in the value of the first ball either changes the parity or not, with equal chances. As a result, there is a $1/2$ chance that the sum of all $n$ balls will be even and $1/2$ that it will be odd.
(This has practical applications in creating unbreakable ciphers, for instance.)