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The numbers $$2$$, $$4$$, $$6$$, $$8$$, $$9$$ and $$11$$ are divided into groups of one or more numbers. The sum of the numbers in each group is the same. Which number must be in the group with the number $$11$$? 
Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
$2+4+6+8+9+11=40$  $2$ groups: sum of each group is $20$, $20=9+11=2+4+6+8$ 