The problem is $n$ fisherman, $n-1$ days.
We need even $n$ for this to make sense, the following construction works: Take a regular polygon with $n-1$ sides. Then pick one color for each of the $n-1$ sides of the polygon, color each diagonal with the color of the side that is parallel to it. Then every vertex will have exactly one diagonal of each color coming out of it, except for the color of the edge opposite to it (it will have no diagonals of this color). So now take an $n$'th vertex and color the line from the new vertex to every old vertex with the unique color the old vertex is missing. This gives you a solution. To see this, associate every fisherman with a vertex, and every color with one of the days.
Here is a drawing of the construction:
![enter image description here](