The sought minimum $m$ is $m=3$, and with what follows, it will be quite evident that the result remains true for $n \geq 3$ lines.
First, we note that, in the obtained configuration, there is at least one non-subdivided region that is a triangle. Indeed, three non-concurrent lines and two never parallel lines form a triangle. Now, any line that crosses a triangle divides it into two polygons, at least one of which is a triangle. Therefore, starting from three of the given lines and "adding" the other 2014 one by one, we are assured of the existence of such a triangular region in the final configuration. The three vertices of such a triangle must be of different colors, which implies that $m \geq 3$.
To conclude, we will construct an appropriate three-coloring. We start by noting that, since there are only a finite number of intersection points, we can choose an orthogonal coordinate system in which these points have distinct x-coordinates. We then number the points $M_{1}, M_{2}, \ldots, M_{k}$ according to increasing x-coordinates (where $k=\frac{2017 \times 2016}{2}$). For any $i$, the point $M_{i}$ has at most four neighbors, and if it belongs to a segment joining two of its neighbors, only one of these two neighbors has an x-coordinate less than that of $M_{i}$. This ensures that, for any $i \leq k$, among the neighbors of $M_{i}$, at most two have indices less than $i$. We can then color the $M_{i}$ in the order of the numbering according to the following procedure: we color $M_{1}$ green and $M_{2}$ red. And, for any $i$ such that $2 \leq i \leq k-1$, if we assume that the points $M_{1}, M_{2}, \ldots, M_{i}$ have been colored either green, red, or blue so that no two adjacent points are (yet) the same color, then, according to the previous remark, at most two neighbors of $M_{i+1}$ have already been colored, leaving one color free for $M_{i+1}$.