The problem belongs to the Czech and Slovak Republic Olympiad teams for Math. I cannot understand the question. Can you explain the question in simple terms?
Show that from any fourteen distinct natural numbers, there exists $k \in \{ {1..7} \}$ for which one can find disjoint k-element subsets $$ {a_{1}, .... ,a_{k}}$$ and $${b_{1}, .... ,b_{k}} $$
of the fourteen numbers such that the sums
$ A= \frac{1}{a_{1}} + ... \frac{1}{a_{k}} , B= \frac{1}{b_{1}} + ... \frac{1}{b_{k}} .$
differ less than 0.001
Solution
Consider the $C(14,7) = 3432$ 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 + \frac{1}{2} + ... \frac{1}{7} $. $1 + 1/2 + ... 1/7 = 363/140 < 2.60$, so each of the 3432 sums lies in one of the 2600 intervals $(0,1/1000], (1/1000,2/1000], ..., (2599/1000,2600/1000]$. 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 $B$.
I assume that the $>$ signs are typos.
You have a set of 14 distinct natural numbers, you can select some of them to put in set $A$ and some other to put in set $B$. There can be anywhere from 1 to 7 elements in each set, but the two sets must have the same number of elements. Some numbers may be in neither set. The sets have to be disjoint (each number can only be in one set). Your goal is to select these sets such that the sum shown in the equation differs by less than 0.001. In other words, the sum of their reciprocal should be less than 0.001.
The question asks you to prove that you can always make such selection, no matter what 14 distinct natural numbers are given to you.