Suppose that $n = 99$. We claim that the answer would be $\frac{4}{3}n$.

First of all, consider these three lines:
- Lines of type $R$: they are vertices drawn from the center;
- Lines of type $L$: they are the largest diagonals;
- Lines of type $S$: the second largest diagonals.

It is clear that the only possible case that these three types have the same color would be $(L, S)$ or $(L, R)$ or $(R, S, S)$. Since every two largest diagonals and every two radii will cut each other, and among each three diagonals of the second largest length, there would be two that cut each other. If we have two lines $L, S$ such that don't cut each other, each radius will cut them.

Now, assign 2 to type $L$ and assign 1 to the types $R, S$. Then, the sum of the weights of these lines is $4n$. On the other hand, each color will cover the number 3, thus, we at least need $\frac{4n}{3}$ colors.

In order to prove this amount is sufficient, we opt to use 4 colors to color all the lines parallel to the three consecutive main diagonals such that the three radii connecting the starting points of these diagonals be of the same color. In the following figure, we present one way to do so!

By rotating this pattern, we shall color the next three consecutive main diagonals and continue this procedure to color all the lines. Since the distance between any two points of the largest distance is $3k+1$. By the time we reach to the other side of the main diagonal, the second main diagonal that starts from that point would be colored and this procedure will continue to color everything. ■