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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.