Solution :

Consider the numbers  from the set  such that  but . We define the first group as  and call  an “extra” number. By the same way, we consider numbers  such that  but  and we define the second group  and call  an “extra” number. We continue this process and define 8 groups and 8 “extra” numbers. Since the sum of all “extra” numbers and all numbers from 8 groups exceeds 15×8 = 120 the sum of all remaining numbers is less than 135 – 120 = 15 and we put all remaining numbers into one group. Finally 8 “extra” numbers we divide into two groups each including 4 “extra” numbers (4×3.5 < 15) and get 8 + 1 + 2 = 11 groups.