Solution. In $N$ weighings, we weigh all $2 N$ stones pairwise. The stones that turned out to be heavier are placed in one pile, and those that are lighter are placed in another. In the first pile, the heaviest stone is found as follows: we number the stones from 1 to $N$, take stones 1 and 2, weigh them, choose the stone with the greatest weight, and compare its weight with stone 3, and so on. On the $(N-1)$-th weighing, the weight of the $N$-th stone is compared with the heaviest of the previous $N-1$ stones. Thus, the heaviest stone in the first pile is guaranteed to be found in $N-1$ weighings. Similarly, we find the lightest stone in the second pile. In total, it will take $N+N-1+N-1=3 N-2$.

Remark. It is possible that students should lose 1-2 points for the lack of a description of the algorithm for selecting the lightest (heaviest) stone.
3. How to find the lightest and heaviest stones among $2 N$ stones, any two of which differ in weight, in $3 N-2$ weighings? All weighings are performed on a two-pan balance without weights.