SELECTION EXAMINATION

(α) If that is possible, for $n=2014$, then for the sum of the elements of $\Sigma$ we get $S_{\Sigma }=S_A+S_B+S_{\Gamma }=3\cdot S_A$, that is, $S_{\Sigma }$ is a multiple of $3$. But $$S_{2014}=1007\cdot 2015\ne \text{multiple of }3.$$ Hence we cannot apply the wanted partition.

(β) For $n=2015$ we have: $S_{2015}=1008\cdot 2015\equiv 0(\mathrm{mod}3)$. We observe that $\Sigma$ consists of the set $M_0=\{1,2,3,4,5\}$ and from $335$ successive six-folds of the form: $$M_k=\{6k,6k+1,6k+2,6k+3,6k+4,6k+5\},k=1,2,\dots ,335.$$ We partition the set $M_0$ into three subsets with equal sums of their elements $A_0=\{1,4\}$, $B_0=\{2,3\}$ and ${\Gamma }_0=\{5\}$. Taking in mind that $$(6k+1)+(6k+4)=(6k+2)+(6k+3)=6k+(6k+5),\text{ for all }k=1,2,\dots ,335,$$ The wanted partition is feasible by taking the sets: $$A=\{1,4\}\cup \{6k+1,6k+4:k=1,2,\dots ,335\}$$ $$B=\{2,3\}\cup \{6k+2,6k+3:k=1,2,\dots ,335\}$$ $$\Gamma =\{5\}\cup \{6k,6k+5:k=1,2,\dots ,335\}.$$

(γ) For $n=2018$ we have: $S_{2018}=1009\cdot 2019\equiv 0(\mathrm{mod}3)$. As in (β) we observe that $\Sigma$ consists of $M_0=\{1,2,3,4,5,6,7,8\}$ and from $335$ successive six-folds of the form: $$M_k=\{6k+3,6k+4,6k+5,6k+6,6k+7,6k+8\},k=1,2,\dots ,335.$$ First we partition $M_0$ into three subsets with equal sums of their elements $A_0=\{1,2,3,6\}$, $B_0=\{5,7\}$ and ${\Gamma }_0=\{4,8\}$. Since for all $k=0,1,\dots ,335$ we have $(6k+3)+(6k+8)=(6k+4)+(6k+7)=(6k+5)+(6k+6)$, the wanted partition is possible by taking $$A=\{1,2,3,6\}\cup \{6k+3,6k+8:k=0,1,2,\dots ,335\}$$ $$B=\{5,7\}\cup \{6k+4,6k+7:k=0,1,2,\dots ,335\}$$ $$\Gamma =\{4,8\}\cup \{6k+5,6k+6:k=0,1,2,\dots ,335\}.$$