Solution:

a) Let $N$ be a box that contains a maximal number of apples, say $n$ pieces. It is first set aside. We now consider all possibilities of dividing the remaining $2004 = 3 \cdot 668$ boxes into three piles, $A$, $B$ and $C$, of 668 boxes each. Let the total number of apples in these boxes be $a$, $b$, $c$. By starting with an arbitrary such division and renaming the piles formed if necessary, we see that there are divisions with $a \leq b \leq c$. Among all of these we fix, for the remainder of the proof, one for which $c - a$ is minimal. If now $c - a > n$, we could swap a box from $A$ containing the maximal number of apples with a box from $C$ containing the minimal number, and, due to the maximality of $n$, after possibly renaming the piles, we would obtain a contradiction to the minimality of $c - a$. Hence $c - a \leq n$, so $a + n \geq c \geq b$, from which immediately follows
$$a + n = \frac{(a + n) + (a + n) + a + n}{3} \geq \frac{c + b + a + n}{3}$$
So if we finally decide to take the 669 boxes $A \cup \{N\}$, i.e. the boxes of pile $A$ together with the box $N$, then we would have at least satisfied the condition on the number of apples to be taken. This naturally holds all the more for $B \cup \{N\}$ and $C \cup \{N\}$. Now let (by the pigeonhole principle) $T$ be one of the three piles that contains at least one third of the pears occurring in them altogether. Thus the selection $T \cup \{N\}$ certainly satisfies the requirement on the number of pears to be chosen and is consequently, by the above, as desired.

b) Here it suffices to give a counterexample. Let one of the boxes contain nothing but one apple, another just one pear, and in each of the remaining 2003 boxes let there lie one peach. If one wanted to satisfy the stated conditions, one would have to choose the box with the apple, the box with the pear, and 668 boxes with a peach, thus needing 670 boxes in total.