Let $A=\{1,2,...,9\}$. We are going to find the number of subsets of $A$ with no four distinct elements satisfying $a+b=c+d$. Let $N$ be a positive integer such that it is possible to write it as sum of at least two pairs of elements of $A$. Then $N$ ranges from $5$ to $15$. One may use the inclusion-exclusion principle to find the number of desired subsets. However, this method is too long and I think there would be an easier and faster way to do the problem.
please give a hint to find a better way. Thank you very much.
No complete answer, but this should help:
As soon as you have two pairs of successive numbers, the equation can clearly be made to work. So, there is no subset of size $7$ or larger for which there are no such $4$ elements.
You can also rule out subsets of size $6$. The only way to not get $2$ distinct pairs of successive numbers is to have $1,3,5,7,9$ plus one more number, but now obviously $1+7=3+5$
So, you 'only' need to check subsets of sizes $4$ and $5$