a) Let $N$ be a box containing a maximum number of apples, say $n$ pieces. It is set aside. We now consider all possibilities of dividing the remaining $2004 = 3 \cdot 668$ boxes into three piles, $A, B$, and $C$, each with 668 boxes. Let the total number of apples in these boxes be $a, b, c$. By starting with any such division and renaming the formed piles if necessary, we see that there are divisions with $a \leqq b \leqq c$. Among all these, we fix for the rest of the proof a division for which $c-a$ is minimal. If $c-a > n$, we could swap a box from $A$ with the maximum number of apples with a box from $C$ with the minimum number of apples and, due to the maximality of $n$, after possibly renaming the piles, we would obtain a contradiction to the minimality of $c-a$. Therefore, $c-a \leqq n$, so $a+n \geqq c \geqq b$, from which it immediately follows that

$$a+n = \frac{(a+n) + (a+n) + a+n}{3} \geqq \frac{c+b+a+n}{3}$$

If we ultimately decide to take the 669 boxes $A \cup \{N\}$, i.e., the boxes of pile $A$ together with the box $N$, we would have at least met the condition on the number of apples to be taken. This is of course also true for $B \cup \{N\}$ and $C \cup \{N\}$. Now, by the pigeonhole principle, let $T$ be one of these three piles that contains at least one-third of the pears occurring in them. Thus, the selection $T \cup \{N\}$ certainly meets the requirement for the number of pears to be chosen and is therefore as desired.
b) Here it suffices to provide a counterexample. Let one of the boxes contain nothing but an apple, another only a pear, and in the remaining 2003 boxes, let there be a peach each. If we wanted to meet the given conditions, we would have to select the box with the apple, the box with the pear, and 668 boxes with a peach, thus requiring a total of 670 boxes.
Remark. In part (a), it is instructive to consider the case where there are 1002 boxes each containing $2m$ apples but no pears, 1002 boxes each containing $2m$ pears but no apples, and one box, $(H)$, containing $m$ apples and $m$ pears, where $m$ stands for a possibly large natural number (e.g., 100000). One can easily convince oneself that box $H$ must be taken. This is still the case even if the described configuration is slightly modified, i.e., by changing the number of fruits (e.g., 1, 2, 3) in comparison to $m$. Most proposed algorithmic approaches of the form "Take boxes with the maximum number of fruits/apples/pears until ... and then always take boxes with the maximum number of ..." do not always lead to taking box $H$ in the described situations. Therefore, they do not solve the problem without further ado.