(WLOG) Ignore all of the blue streets. Then the white and red streets form a series of alternating edge-colored polygons, as each intersection must have one white and one red street and the streets cannot cross. Also, the polygons are non-intersecting, and they each have an even number of edges. Consider all polygons $ P$ which are not enclosed by any polygons, and going clockwise around the vertices, alternately label them $ A$ and $ B$, such that (wlog) each $ A$ is between a white and red street in that order. Repeat this step on any polygons enclosed by the above polygons, except reverse the $ A$s and $ B$s such that each $ A$ is now between a red and white street in that order, and so forth continue labeling all of the vertices, alternating the order with each 'level'. It is easy to see now that if 2 $ A$s or 2 $ B$s are connected by a blue street, then they have the same sign (consider if the blue path is within a polygon, between two polygons, or between a polygon and the polygon enclosing it), and if an $ A$ and $ B$ are connected they have opposite signs. We ignore the case with opposite signs because they do not change the difference. There are the same number of $ A$s and $ B$s, and after removing the pairs of points with opposite signs, there still are the same number of $ A$s and $ B$s. Since the streets connect either 2 $ A$s or 2 $ B$s, then there are $ 2n$ of each type, with $ 4n$ total intersections. But the number of positive and negative intersections are even, so we are done.