I solved it when I was on my way from the home to the school. Suppose a contrary. Consider the bottom side of the triangle. It contains 3001 vertices. So there are at least 1001 points of the same color (color 1). Consider all possible triangles, whose two 2 vertices coincide with two of these points. There are at least $1001\cdot 1000/2=500500$ such triangles. Their third vertices of color 2 or 3, and they are placed on 3000 horizontal lines, so there is a horizontal line, which contains at least 167 of these vertices. At least 84 have the same color (color 2). If we consider a triangle, whose two vertices coincide with two of these 84 vertices, then third vertex is colored (!) in color 3. There is at least $84\cdot 83/2=3486$ such triangles, so there is two vertices of color 3 on the same horizontal line, but then third vertex for the corresponding triangle cannot be colored in color 1, 2 or 3. Contradiction.