I claim that the answer is $2015$: If we represent the problem using graph theory, we need the minimal possible number of colors to color the edges of the complete graph $K_{2015}$ such that every triangle (or $3$-cycle) has edges of different colors. First let us prove that we need at least $2014$ colors. Indeed, take an arbitary vertex. It has degree $2014$. If we would have less than $2014$ colors, then by pigeonhole principle two edges coming from that vertex would have the same color, so we would have a triangle with two same colored edges. We can also see that $2014$ colors are not enough. Indeed, the graph has $\binom{2015}{2}$ edges. If $2014$ colors were enough, then by pigeonhole principle we would have at least $\left[ \frac{2015 \cdot 2014}{2 \cdot 2014} \right ] + 1 = 1008$ edges of a certain color. But $2\cdot1008 = 2016 > 2015$ so at least two of them would share a vertex, which would give a triangle with at least two same colored edges. Now let's prove that $2015$ colors is enough. Numerate the vertices with $1,2,...2015$ in some order. Color the edge that connects the vertices $i$ and $j$ with color $i + j \; mod \; 2015$. If a triangle would have two same colored edges, then they must share a vertex $a$. Let the other two vertices of the triangle be $b$ and $c$. But then we would have $a+b \equiv a+c \; mod \; 2015 \leftrightarrow b \equiv c \; mod \; 2015$ which is impossible because $b \neq c$ and $2015 \nmid b-c$. We conclude that the construction indeed works.