In a country consisting of $2015$ cities, between any two cities there is exactly one direct round flight operated by some air company. Find the minimal possible number of air companies if direct flights between any three cities are operated by three different air companies.
1. **Graph Representation and Problem Restatement**:
   - Represent the problem using graph theory. We need to find the minimal 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.

2. **Lower Bound on the Number of Colors**:
   - Consider an arbitrary vertex in \( K_{2015} \). This vertex has a degree of \( 2014 \) because it is connected to every other vertex.
   - If we use fewer than \( 2014 \) colors, by the pigeonhole principle, at least two edges incident to this vertex would have the same color. This would form a triangle with two edges of the same color, violating the problem's condition.
   - Therefore, we need at least \( 2014 \) colors.

3. **Insufficiency of 2014 Colors**:
   - The complete graph \( K_{2015} \) has \( \binom{2015}{2} = \frac{2015 \cdot 2014}{2} \) edges.
   - If \( 2014 \) colors were enough, then by the pigeonhole principle, at least one color would be used for at least \( \left\lceil \frac{2015 \cdot 2014}{2014} \right\rceil = 1008 \) edges.
   - However, if a single color is used for \( 1008 \) edges, then by the pigeonhole principle, at least two of these edges would share a vertex, forming a triangle with two edges of the same color.
   - This contradiction shows that \( 2014 \) colors are not sufficient.

4. **Sufficiency of 2015 Colors**:
   - We now show that \( 2015 \) colors are sufficient.
   - Number the vertices \( 1, 2, \ldots, 2015 \).
   - Color the edge connecting vertices \( i \) and \( j \) with color \( (i + j) \mod 2015 \).
   - Suppose a triangle has two edges of the same color. Let the vertices of the triangle be \( a, b, \) and \( c \).
   - If the edges \( (a, b) \) and \( (a, c) \) have the same color, then \( (a + b) \mod 2015 = (a + c) \mod 2015 \).
   - This implies \( b \equiv c \mod 2015 \), which is impossible because \( b \neq c \) and \( 2015 \nmid (b - c) \).
   - Therefore, no triangle can have two edges of the same color, and \( 2015 \) colors are sufficient.

The final answer is \( \boxed{2015} \).