, 2015

Let $a_0 < a_1 < \dots < a_{n-1}$ be elements of $X$. For each positive integers $i < n-m$ and $j \leq m$, construct a subset $A_{ij}$ of $X$ as $A_{ij} = \{a_i, \dots, a_{i+m}\} \setminus \{a_{i+j}\}$. Then it is clear that $S(A_{ij}) \in X_m$ and
$$S(A_{i0}) > S(A_{i1}) > \dots > S(A_{im-1}) > S(A_{im}). \qquad (1)$$
Moreover, we have
$$X_m = \{S(A_{ij}) : 0 \leq j \leq m \text{ and } 0 \leq i < n-m\},$$
since $S(A_{i0}) = S(A_{i+1m})$. For each $i < n - m - 1$ and $j < m$, we consider the subset $B_{ij}$ of $X$ given by $B_{ij} = A_{ij} \cup \{a_{i+m+1}\} \setminus \{a_{i+m}\}$. Certainly, $S(B_{ij}) \in X_m$ and
$$S(A_{i+1m-1}) = S(B_{i0}) > S(B_{i1}) > \dots > S(B_{im-1}) > S(A_{im-1}).$$
Thus, between $S(A_{im-1})$ and $S(A_{i+1m-1})$, there are $m-1$ different elements of $X_m$ and hence $S(A_{ij}) = S(B_{ij+1})$ from (1). Therefore we get that
$$a_{i+j+1} + a_{i+m} = a_{i+j} + a_{i+m+1},$$
which implies that $a_0, a_1, \dots, a_{n-1}$ are in arithmetic progression.