consider the arithmetic sequence u_p=1+3(p-1) we have u_n - u_p = 3(n-p) u_n + u_p = 2 + 3(n+p-2) We can see that the difference cannot divide the sum. If we put any other number in the sequence of the form 3q or 3q+2 we can form a difference equal to 1 which will divide the sum. So i think that tha max number is (n- r)/3 where r is the remainder of n mod3

senouy's argument almost hits the spot. We partition $ A$ into subsets with three consecutive intergers: $ \{1,2,3\}$, $ \{4,5,6\}$, $ \ldots$; therefore, we get $ m: =\left\lceil\frac{n}{3}\right\rceil$ subsets. Call this partition $ \mathcal{P}$. If we have more than $ m$ elements in $ B$, then at least two elements of $ B$ must lie in the same subset within $ \mathcal{P}$, and this violates the requirement. So, $ B$ has at most $ m$ elements. An example which $ |B|=m$ is easy to find: $ B=\{1,4,7,\ldots\} \cap A$.