TITLE: Number theory/combinatorics problem?
QUESTION [6 upvotes]: Given any set of $14$ (different) natural numbers, prove that for some $k$ ($1 ≤ k ≤ 7$)
there exist two disjoint $k$-element subsets $\{a_1,...,a_k\}$ and $\{b_1,...,b_k\}$ such that the sum of the reciprocals of all elements in each set differ by less than $0.001$, i.e. $|A−B| < 0.001$, where $A =$ the sum of the reciprocals of the elements of the first subset and $B =$ the sum of the reciprocals of the elements of the second subset.
Note: this problem is from the $1998$ Czech and Slovak Math Olympiad

REPLY [13 votes]: Consider the $\binom{14}{7} = 3432$ $7$-element subsets of the fourteen numbers, and look at the sums of the reciprocals of the numbers in each subset. Each sum is at most$$1 + {1\over2} + \ldots + {1\over7} = {{363}\over{140}} < 2.60,$$so each of the $3432$ sums lies in one of the $2600$ intervals$$\left(0, {1\over{1000}}\right], \text{ }\left({1\over{1000}}, {2\over{1000}}\right], \text{ }\ldots\text{ }, \text{ }\left({{2599}\over{1000}}, {{2600}\over{1000}}\right].$$Then by the pigeonhole principle, some two sums lie in the same interval. Taking the corresponding sets and discarding any possible common elements, we obtain two satisfactory subsets $A$ and $B$.