$$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10+11+12 = 78$$, $$80 -- 78 = 2$$. Since all the numbers should be different, the remaining \textquotesingle$$2$$\textquotesingle{} cannot make up a new positive integer. It can only be added to the number(s) before. Therefore, there are at most $$12$$ different positive integers that can add up to $$80$$. 
How many different positive integers at most can add up to $$80$$? 
- $$9$$ 
- $$10$$ 
- $$11$$ 
- $$12$$ 
- $$13$$ 