Assume the height of every little triangle is $1$.
Define the a,b,c-coordinate of a triangle as the minimum distance from a point on its boundary to the top side, the left-bottom side, the right-bottom side of the hexagon, respectively.
The sum of the three coordinates of a black triangle is $3n-1$, while the sum of a white triangle is $3n-2$.
The $2n$ triangles chosen has a-coordinate from $0$ to $2n-1$, so the sum of them is $2n^2-n$(b,c-coordinates, too). The sum of all the coordinates of the chosen triangle up(3 coordinates and $2n$ triangles) is $6n^2-3n$. If there are $x$ black triangles and $y$ white triangles chosen, then the sum of all coordinates can be computed again as $(3n-1)x+(3n-2)y$ with $x+y=2n$.
Hence $6n^2-3n=3n(x+y)-x-2y, x+2y=3n$
We have $x=y=n$.
