Take $N-a_i = b_i^2$ for $1\leq i \leq 6$. Since $a_i$ are required to be distinct, so should be $b_i$. By summation over $i$ we then get $5N = \sum_{i=1}^6 b_i^2$. The least eligible value is for $b_i$ being $0,1,2,3,4,5$, whence $N = 11$ and $a_i$ are $11,10,7,2,-5,-14$.

Great job! You solved it so quickly! It takes me 1.5 hours to solve it...

I' m sorry I think I mistype the problem, $ a_1, a_2, a_3, a_4, a_5, a_6 $ all must be greater than 0... The right problem is: Determine the smallest positive integer $ N $ such that there exists 6 distinct integers $ a_1, a_2, a_3, a_4, a_5, a_6$ $ [color=\#BF0080]> 0[/color] $satisfying: (i) $ \ N = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 $ (ii) $ \ N - a_i $ is a perfect square for $ \ i = 1,2,3,4,5,6 $.

http://www.artofproblemsolving.com/Forum/resources.php?c=150&cid=151&sid=9a0863dfbecea64e0cddb29f01754f52 here Problems from 2009 are not posted.Will our admins move this to that section?

The answer is $ 911 $, with $ b_i = (25,26,27,28,29,30) $ and $ a_i = (286, 235, 182, 127, 70, 11) $