I think you have a small error in typo. Although,it is a well-known problem of Erdos and Suranyi.You can see more in here: http://www.mathlinks.ro/Forum/viewtopic.php?t=30998

I'm sorry, it should be $ a_{1}\cdot1^2 + a_{2}\cdot2^2 + \cdots + a_{n}\cdot n^2 = 0.$

No Reason wrote: I think you have a small error in typo. Although,it is a well-known problem of Erdos and Suranyi.You can see more in here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30998 I don't think it's a same problem. Here is my solution: Obviously $n \equiv 3, 0 \pmod4 $. It cannot happen $n=3$ or $n=4$. Let $n \ge 7$. Now: $n=7$ works: $1^2+2^2+4^2+7^2=3^2+5^2+6^2$ $n=8$ works: $1^2+4^2+6^2+7^2=2^2+3^2+5^2+8^2$ $n=11$ works: $11^2+9^2+7^2=rest$ $n=12$ works: $12^2+10^2+9^2=rest$ Now if $n$ works then also $n+8$ works because: $(n+1)^2+(n+4)^2+(n+6)^2+(n+7)^2=(n+2)^2+(n+3)^2+(n+5)^2+(n+8)^2$. End of story.