I suppose the elements in A are positive integers, right? Let the smallest element in A be n. Then we partition the numbers from n+1 to 101 as follows: {n+1, 2n+1}, {n+2, 2n+2}, …, {2n, 3n} {3n+1, 4n+1}, {3n+2, 4n+2}, …, {4n, 5n}, … Singletons are allowed. Pairs as formed until both elements of some pairs exceed 101. Note that the numbers in each pair differ by n.

Let for some and , then it is easy to show that there are pairs where . We can show that . Therefore, . Therefore, there are not more than 50 pairs.

If 52 numbers are chosen, there 51 of them must be chosen from at most 50 pairs. By Pigeonhole Principle, two of them must be from the same pair. As the numbers in the pair differ by n but n is also chosen, a contradiction occur. From the construction method, we should see that there are some ways to pick 51 of them without contradicting the given condition. For example, {1, 3, 5, …, 99, 101} and {51, 52, 53, …, 101} are two ways to do so._________________

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