I claim that $ r = 200. $ For all $ k \in \{1, 2, \dots, 10\} $ let $ A_k = \{100k - 99, 100k - 98, \dots, 100k\}. $ Then if we look at the subsets $ A_1 \cup A_5 \cup A_6 \cup A_7 \cup A_9 $ and $ A_1 \cup A_2 \cup A_7 \cup A_8 \cup A_{10} $ and $ A_2 \cup A_3 \cup A_6 \cup A_8 \cup A_{9} $ and $ A_3 \cup A_4 \cup A_7 \cup A_9 \cup A_{10} $ and $ A_4 \cup A_5 \cup A_6 \cup A_8 \cup A_{10} $ we see that any two of these subsets share exactly $ 200 $ elements which implies that $ r \le 200. $ Consider any five subsets $ S_1, S_2, S_3, S_4, S_5 $ of $ \{1, 2, \dots,, 1000\} $ each with $ 500 $ elements. Let $ a_i $ denote the number of subsets the integer $ i $ is a part of. It is clear that $ \sum_{i = 1}^{1000}a_i = 2500. $ It's also clear that $ \sum_{1 \ge i < j \le 5}|S_i \cap S_j| = \sum_{i = 1}^{1000}\binom{a_i}{2}. $ By convexity (or smoothing, take your pick) we find that $ \sum_{i = 1}^{1000}\binom{a_i}{2} \le 500\binom{3}{2} + 500\binom{2}{2} = 2000 $ so by the pigeonhole principle there exists distinct $ i, j \in \{1, 2, 3, 4, 5\} $ such that $ |S_i \cap S_j| \ge \frac{2000}{\binom{5}{2}} = 200 $ so $ r \ge 200. $ Therefore $ \boxed{r = 200} $ as desired.

This problem is just the discrete counterpart to the "beggar's coat" problem from the famed Arthur Engel book (a beggar's coat has five patches, each of area half that of the coat; prove two of the patches overlap on at least an area of one fifth that of the coat). The method(s) of proof are obviously identical. The problem was first submitted (in a slightly different version) as a proposal for the RMM (and was rightfully so dismissed). Pouah ...

lemma: $\binom{2a}{2} + \binom{2b}{2} > 2.\binom{a+b}{2}$ let those sets :$A_1 ,... ,A5$ for example if number $t$ exist $A_1,A_3$ replace $t$ with set ${A_1,A_3}$ and let $a_t$ be the number of elements on this set . for example in this set $a_t=2$ so obviousely $r$ should less than : $M=$${\sum_{i = 1}^{1000}\binom{a_i}{2}}/10 $ with this assumption $ \sum_{i = 1}^{1000}a_i = 2500. $. according lemma easy understanding minimum of $M$ is $500\binom{3}{2} + 500\binom{2}{2}/10 = 2000/10=200 $

Wolstenholme wrote: $ \sum_{i = 1}^{1000}\binom{a_i}{2} \le 500\binom{3}{2} + 500\binom{2}{2} = 2000 $ Quote: Wolstenholme your proof is ok but im not sure about this part i think you should change < with >

arzhang2001 wrote: Wolstenholme wrote: $ \sum_{i = 1}^{1000}\binom{a_i}{2} \le 500\binom{3}{2} + 500\binom{2}{2} = 2000 $ Wolstenholme your proof is ok but im not sure about this part i think you should change < with >