It is quite similar to my combinatorics problem, but I couldn't have solved it. I hope that someone can solve this problem, so I probably can solve my problem. It's here: https://artofproblemsolving.com/community/c6h2629419_a_short_but_amazing_problem_about_combinatorics

parmenides51 wrote: Find the least possible number of elements which can be deleted from the set $\{1,2,...,20\}$ so that the sum of no two different remaining numbers is not a perfect square. N. Sedrakian , I.Voronovich Firstly the maximum sum is 38 so we are considering squares less than 38. Now it is just casework:- 1)$a+b=36$ Here,atleast 2 of $(18,18),(19,17)$ should be deleted. 2)$a+b=25$ Here,atleast 1 from each set of $(19,6),(18,7),(17,8),(16,9),(15,10),(14,11),(13,12)$ should be deleted. 3)$a+b=16$ Here atleast one of $(15,1),(14,2),(13,3),(12,4),(11,5),(10,6),(9,7),(8,8)$ should be deleted 4)$a+b=9$ Here,atleast 1 from each set of $(8,1),(7,2),(6,3),(5,4)$ should be deleted. 5)$a+b=4$ Here,atleast 1 from each set of $(3,1),(2,2)$ should be deleted. Now $2,8,18$ are nessecary to be deleted.now either 1 or 3 to be removed so 1 more number. Now from the second one another 1 numbers to be removed.Now a little more checking will give $\boxed{min(\text{Numbers to be deleted})=11}$