Balkan Mathematical Olympiad Shortlist

We will prove that this is possible. Total number of subsets of the set $M$ is $2^{2013}$. On circle $\Gamma$ we arbitrarily choose $2^{2013}$ points which are vertices of a regular $2^{2013}$-gon. We join subsets of the set $M$ and chosen points in the following manner: if we join subset $\mathcal{A}$ with some point on $\Gamma$, we join subset ${\mathcal{A}}^c=M\setminus \mathcal{A}$ with a point symmetric to the point joined with $\mathcal{A}$ with respect to the center of $\Gamma$ (number $2^{2013}$ is even, so this is possible). If $A_1,A_2,\dots ,A_k$ are some of the points joined with subsets, which are vertices of a regular $k$-gon, then it follows that $k\mid 2^{2013}$, so $k$ is a number divisible by $4$, say $k=4t$. That is why all points $A_1,A_2,\dots ,A_k$ can be divided in $\frac{k}{2}=2t$ pairs of symmetric points with respect to center of $\Gamma$. Using that $$S(\mathcal{A})+S({\mathcal{A}}^c)=1+2+\cdots +2013=1007\cdot 2013,$$ we get $S({\mathcal{A}}_1)+\cdots +S({\mathcal{A}}_k)=2t\cdot 1007\cdot 2013=2014\cdot 2013\cdot t$, so all conditions are fulfilled.