There are infinitely many numbers less than or equal to n (assuming n is a real number). From this it follows that there are infinitely many sets of un-average numbers, each of whose elements is less than or equal to n. So there is no "largest" such set. For any such set, we could always add numbers in such a way that the average of all of the numbers is not a member of the set.

Edward A.
Kevin’s point is well-taken: if the numbers can be reals, the question is degenerate and too easy. Let’s suppose the problem is intended to be more interesting. Let’s restrict it to positive integers. Here is a set of size 5 that is un-average: {1,2,6,7,9} with numbers less than or equal to 10. also {1,2,4,5,10}. How can we generate un-average sets? Here’s an observation: If n is in a set, the set cannot contain both n+m and n+ 2m, for any m. To be concrete, a set of values <= 10 must obey the following constraints: each set of constraints relates to one n, showing each forbidden pair (n+m, n+2m), for all m such that n+2m <= 10. 1: (2,3)/(3,5)/(4,7)/(5,9) 2: (3,4)/(4,6)/(5,8)/(6,10) 3: (4,5)/(5,7)/(6,8) 4: (5,6)/(6,8)/(7,10) 5: (6,7)/(7,9) 6: (7,8)/(8,10) 7: (8,9) 8: (9,10) This observation gives a relatively simple way to sieve candidate un-average sets, but it does not give a way to establish the largest numbered set, except by enumeration. I believe I have found that for integers in [1,10], the largest set size is 5. I have not extrapolated that to [1-20]. Naz, this is an interesting problem, but I can’t offer much help in answering it.09/25/19

Kevin C.
09/25/19

Kevin C.
09/25/19

Kevin C.
09/25/19
Naz N.
can you provide me with some ways for figuring out the numbers ? or any examples that can help me understand it even more09/23/19