Romanian Mathematical Olympiad

An integer $n$ belongs to $M$ if and only if there exists disjoint subsets $A$, $B$ of the set $S = \{1, 2, \dots, 2015\}$, with $A \cup B = S$, so that $-2a + 3b = n$, where $a$ is the sum of the elements of $A$ and $b$ is the sum of the elements of $B$ (the sum of the elements of the empty set being $0$).

Then $n = 5b - 2(a+b)$ and, since $a+b = 1+2+\dots+2015$, it follows that
$n = 5b - 2015 \cdot 2016$.

This shows that $n$ is divisible by $5$, hence $2016 \notin M$.

In order to prove that $2015 \in M$, it is enough to find $B \subset \{1, 2, \dots, 2015\}$ with the sum of its elements $b = \frac{1}{5}(2015 \cdot 2016 + 2015) = 403 \cdot 2017$.

An example is when $B$ is the union of $403$ pairs of elements of $S$ with sum $2017$, for instance $(2, 2015)$, $(3, 2014)$, ..., $(404, 1613)$.

So, taking $x_2 = x_3 = \dots = x_{404} = x_{1613} = x_{1614} = \dots = x_{2015} = 3$ and $x_1 = x_{405} = x_{406} = \dots = x_{1612} = -2$, we get $2015 \in M$.