What's the answer?

N = 2 × 100 – 1 = 199 ?

The answer is $\boxed{205}$. The key is to use the graphical interpretation and notice that in each connected component we can determine which boxes belong to different colors but can not determine which color is which. Construction: Write the pairs $(1,2),(1,3),\dots,(1,206)$ from which we can determine all the colors. Bound: WLOG assume the graph has three connected components adding edges if necessary. Case 1: The graph contains a connected component of even size with size at most $200$ It is possible for half of the elements of this component to be of different color making it impossible to come to a conclusion. Case 2: The graph contains a connected component of even size with size greater than $200$ The two other components must have odd size, a contradiction. Case 3: The graph contains three components of odd size Out of the two smallest components each could have just over half of the elements belonging to one color over the other making it impossible to determine the color of the boxes.

If box 1 is red, then at the max I can make a red pair with 99 other boxes If box 1 is white, then at the max it can make a non-pair with 100 other boxes Therefore, after comparing 199 pairs from 200 boxes (box 1 with all the other 199) , the worst case scenario is having 99 pairs and 100 non-pairs, which is possible to both sides (red 1 vs white 1) Then the 200th comparison will return either a pair or a non-pair, from which we can determine what was in box 1 and relatively the other boxes, too Answer is actually 2 × min(100, 107) = 200 150 comparisons at 151st box we may find 100 P + 50 NP or 49P + 101 NP , but there are 56 more boxes with leftover 49 comparisons. First do 28 comparisons, up to 6 NP. If all 6NP turned up, just do 6 more comparison of one member of the NPs with any one of the Ps (or just the specific colours you need and already found out which from the first 151 boxes) On the other hand, if we get 28 pairs previously, then all the remaining whites are pairing together in 2 × 3 so we do 14 more comparisons, for a guaranteed of 1 NP or exactly 3 NPs Anyway, since our worst case scenario can happen within 99 + 100 = 199 comparisons by exhausting one of the complementary possibilities, that means doing it just once more can resolve the situation

If she writes all pairs down at once then $205$ is right. (weird wording)

How can i find the official solution?