Xixas wrote: Find the maximal cardinality $|S|$ of the subset $S \subset A=\{1, 2, 3, \dots, 9\}$ given that no two sums $a+b | a, b \in S, a \neq b$ are equal. $S=\{1,2,3,5,8\}$ verify If $|S|\ge 6$ there are 15 sums. but $3\le a+b\le 17$ (15 numbers) Therefore the 15 sums are 3,4,...,17 therefore $1,2\in S$ and $8,9\in S$ 1+9=2+8(contradiction) Hence the maximal cardinality is 5

Canadian Olympiad 2002.

The answer is $5$. One possible construction is $S = \{1, 2, 3, 6, 9\}$, which clearly works. Now, since $$1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 = 10$$we know our upper bound is (also) $9 - 4 = 5$, as at most one element from each of the pairs $$\{(1, 9), (2, 8), (3, 7), (4, 6)\}$$can belong to $S$. $\blacksquare$