Let $A\subset\{1,11,21,31,...,541,551\}$ such that no two elements of $A$ add up to $552$. Then show that $A$ cannot have more than $28$ elements.
Let $S=\{1,11,21,31,...,541,551\}$, whence $|S|=56$. Then surely $A\le56$, but how come $A\le\frac{56}{2}$ because of the 2nd restriction in the question? Any hint will be great. Also, is there an equivalent combinatorial presentation of the problem?
Generalising, if $A\subset\{a,a+d,a+2d,...,a+nd\}$ where $a,n,d$ are real numbers and no two elements of $A$ add up to $a+nd+1$, then does $|A|\le\frac{n+1}{2}$ hold?
EDIT.
After browsing some books I found that this particular property of a set has a name. The formal definition goes as,
A set $S$ of integers is called sum-free if $x+y\ne S$ for every $x,y\in S$ ($x,y$ are not necessarily distinct here).
There was this particular exercise in applications of the Pigeonhole principle asking how big a sum-free subset of $\{1,2,...,2n+1\}$ could be, but there was no mention for a general arithmetic progression of the elements. The answer to this question was $n+1$, where the hint given was about the same as the answer already posted here. So I suppose my initial idea of generalising was wrong.
You have that $\frac{1+551}2=276$ and hence you can split $S$ in two sets with $28$ elements each, as follows: $$S_1=\{1,11,\dots,271\}, \quad S_2=\{281,282,\dots,551\}$$ with $28$ elements each. Sort $S_1$ in ascending order and $S_2$ in descending and observe that $$s_1(j)+s_2(j)=552$$ for $j=1,2,\dots,28$. Now, if $A$ has $29$ elements or more, you know that there exists $1\le j\le 28$ such that $$s_1(j),s_2(j)\in A$$