If there are exactly $n$ objects of each colour, we are done easily. Now suppose that this is not true, therefore there is at least one colour (say red) with more than $n$ objects and one colour(say blue) with less than $n$ objects. In your first box place ALL the blue objects and fill the rest of the box up with red objects. Now we have $n(k-1)$ objects left and $k-1$ boxes left and $k-1$ colours left (as we completed got rid of blue), hence we can just use induction.