Let us choose some point $P$ from the initial set. Let $Q_1,Q_2,\hdots,Q_n$ be the points of the initial set different from $P$ and let $0, 1,\hdots , k-1$ be the colors. For each pair of points $A$ and $B$, denote the color assigned to the segment $\overline{AB}$ by $f(A,B)$. Consider a coloring such that $f(Q_i,Q_j)=i+j \bmod k$ and $f(P,Q_i)=2i \bmod k$ for any $Q_i$ and $Q_j$. We will show that this coloring satisfies the conditions of the problem. Notice that, for each $i$, the segments passing through point $Q_i$ have assigned the colors $i+1,i+2,\hdots, 2i,\hdots, i+n \pmod k$. Since there are at least $\left \lfloor \frac{n}{k}\right \rfloor$ integers of the set $\{i+1,i+2,\hdots,i+n\}$ congruent to $m$ modulo $k$, for each $0\leq m\leq k-1$, it follows that there are at least $\left \lfloor \frac{n}{k}\right \rfloor$ segments of each color passing through $Q_i$. Then, given any pair of colors, the number of good angles formed only by segments of those colors is at least $\left \lfloor \frac{n}{k}\right \rfloor^2$. Since there are ${k \choose 2}$ possible pairs of colors, for each $i$ the number of good angles whose vertex is the point $Q_i$ is at least ${k \choose 2} \times {\left \lfloor \frac{n}{k}\right \rfloor}^2$. In particular, since there are $n$ points satisfying the above, the number of good angles is at least $n \times {k \choose 2} \times \left \lfloor \frac{n}{k}\right \rfloor^2$. Finally, to see that the total number of good angles is strictly greater than this number, let's consider the segments $\overline{PQ_1}$ and $\overline{PQ_2}$, these segments have assigned the colors $2\pmod{k}$ and $4\pmod{k}$ respectively, since these colors are different from each other for all $k\geq 3$ we conclude that point $P$ is vertex of at least one good angle and therefore there are at least $$n \times {k \choose 2} \times {\left \lfloor \frac{n}{k}\right \rfloor}^2 +1$$good angles, concluding the problem.