(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
$$S_{2014} = 1007 \cdot 2015 \neq \text{multiple of } 3.$$
Hence we cannot apply the wanted partition.

(β) For $n = 2015$ we have: $S_{2015} = 1008 \cdot 2015 \equiv 0 \pmod{3}$.
We observe that $\Sigma$ consists of the set $M_0 = \{1,2,3,4,5\}$ and from $335$ successive six-folds of the form:
$$M_k = \{6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5\}, \quad k = 1, 2, \dots, 335.$$
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: $S_{2018} = 1009 \cdot 2019 \equiv 0 \pmod{3}$. 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:
$$M_k = \{6k+3,6k+4,6k+5,6k+6,6k+7,6k+8\}, \quad k = 1,2,\dots,335.$$
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
$$A = \{1,2,3,6\} \cup \{6k+3,6k+8:k=0,1,2,\dots,335\}$$
$$B = \{5,7\} \cup \{6k+4,6k+7:k=0,1,2,\dots,335\}$$
$$\Gamma = \{4,8\} \cup \{6k+5,6k+6:k=0,1,2,\dots,335\}.$$