a) The answer is in the negative. If we have 100 boxes, and 25 of the boxes only contain one red ball each, while the other 75 boxes contain just one blue ball each, then one needs to choose at least 13 of the boxes containing red balls and at least 38 of the boxes containing blue balls, which means one would need to choose at least 51 boxes. Thus, such a choice is not always possible.

b) We prove by induction that from $2n + 1$ boxes one can always choose $n + 1$ that contain at least half of the balls of both colors.

For $n = 1$: if choosing two boxes would not be possible, then the remaining box must contain the majority of the red balls, or the majority of the blue balls. But there is only one such choice for each color, and there are 3 ways of choosing the boxes, so at least one of the choices must work.

The inductive step. We say two boxes are comparable if one of them has at least as many red balls as the other one and at least as many blue balls as the other one. If two comparable boxes exist, we eliminate them and apply the inductive hypothesis to the remaining $2n - 1$ boxes. Then, to the $n$ boxes thus chosen, we add that box of the two comparable ones that contains more red and more blue balls. If no two boxes are comparable, we denote by $a_i$ and $b_i$ the number of red balls and blue balls in box no. $i$, respectively. Changing the order of the boxes, one may assume that $a_1 > a_2 > \dots > a_{99}$ and $b_1 < b_2 < \dots < b_{99}$. In this case, simply choose the boxes whose ranks are odd (boxes 1, 3, 5, $\dots,$ 99).

An easier solution to point b) (given in the contest by Dinu Iosifescu). Denoting by $r_i$ the number of red balls in the box $i$, we may assume, without loss of generality, that $r_1 \ge r_2 \ge \dots \ge r_{99}$. Then we can choose box no. 1, then, from box no. 2 and box no. 3 we choose the one containing more blue balls, from box no. 4 and box no. 5 we choose the one containing more blue balls, and so on from box no. 98 and box no. 99 we choose the one containing more blue balls. It is clear that this choice satisfies the conditions in the statement.