67th 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 $a+b=1+2+\cdots +2015$, 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 $x_2=x_3=\cdots =x_{404}=x_{1613}=x_{1614}=\cdots =x_{2015}=3$ and $x_1=x_{405}=x_{406}=\cdots =x_{1612}=-2$, we get $2015\in M$.