Yet another very easy one we chose to ignore . The sum of all the numbers at point (a) is a prime, so that's that. For (b), if $n>6$ or if $n=4$ (the only composite below $6$), then there is a divisor $d>1$ of $n$ which divides the sum of all the elements in the set, which is $S=na^2+n(n+1)ab+\frac{n(n+1)(2n+1)}6b^2$. On the other hand, at least one of $a+b,a+2b,\ldots,a+nb$ is divisible by $n$ and thus by $d$, meaning that it's not coprime with the sum $S$. The answer must be $n=6$.

Not that the problem is not trivial, but could I know dear Grobber why at least one of the n numbers must be a multiple of n?

Sorry about that, I rushed into it, and that's just yet another stupid mistake of mine. If $b$ is not coprime with $\frac n{(n,6)}$, then take $d=(b,\frac n{(n,6)})$. We then have $d|ka^2+k(k+1)ab+\frac{k(k+1)(2k+1)}6b^2=(a+b)^2+\ldots+(a+db)^2$, and we also have $d|$ the sum of all the numbers. On the other hand, if $b$ is coprime with $k=\frac n{(n,6)}$, one of $a+b,a+2b,\ldots,a+kb$ is divisible by $k$. Is it Ok now?

Ah, Grobber! You are so quick. I was writing my correction to you post Thus my work is vain...

Yes, grobber, but you see... obvious things always produce bad things!

Don't blame Grobber to much, Harazi!

Hmm.. I don't know what that means.. I'm sure I'm supposed to learn a lesson from all of this though . Sorry again .