An equivalent statement is : partition the set $\{1,2,...3n\}$ into three disjoint subsets $A=\{a_1,a_2,...a_n\}$, $B=\{b_1,b_2,...b_n\}$, and $C=\{c_1,c_2,...c_n\}$, so that $a_i+b_i=c_i$ . It can be shown that the partition exists for $n=$ power of $4$ or $n=(3^k-1)/2$ for some $k$. The construction is quite tricky, and I have no idea how someone can come up with that.
For example, when $n=4^k$, let $A=\{a_1,a_2,...a_n\},B=\{b_1,b_2,...b_n\},C=\{c_1,c_2,...c_n\}$ be a solution. When $n'=4^{k+1}=4n$, we construct $A'$, $B'$, and $C'$ as follows: $$A'=\{1,3,5,7...2*3n-1,2a_1,2a_2,...2a_n\},$$ i.e, the union of odd numbers up to $2\cdot 3n-1$ and set $$B'=\{9n,9n-1,...9n+1-3n=6n+1,2b_1,2b_2,...2b_n\},$$ and $$C'=\{9n+1,9n+2,...9n+3n=12n,2c_1,2c_2,...2c_n\}.$$
Could anyone please give me some tips on how to construct solutions naturally?