Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

What a problem. Here's my construction for $k=5$ in its full glory. [asy][asy] size(8cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen reddraw,greendraw,bluedraw,purpledraw; reddraw = RGB(255,85,85); greendraw = RGB(0,187,0); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pair A1,A2,A3,A4,A5,A6,A7,A8,A9,A10; A1 = dir(288); A2 = dir(324); A3 = dir(0); A4 = dir(36); A5 = dir(72); A6 = dir(108); A7 = dir(144); A8 = dir(180); A9 = dir(216); A10 = dir(252); draw(A2--A3--A4--cycle,reddraw); draw(A7--A8--A9--cycle,reddraw); draw(A1--A6,reddraw); draw(A5--A10,reddraw); draw(A1--A3--A5--cycle,purpledraw); draw(A6--A8--A10--cycle,purpledraw); draw(A2--A9,purpledraw); draw(A4--A7,purpledraw); draw(A2--A5--A8--cycle,bluedraw); draw(A3--A7--A10--cycle,bluedraw); draw(A1--A9,bluedraw); draw(A4--A6,bluedraw); draw(A1--A4--A8--cycle,greendraw); draw(A3--A6--A9--cycle,greendraw); draw(A2--A10,greendraw); draw(A5--A7,greendraw); draw(A3--A8); draw(A9--A10--A1--A2); draw(A4--A5--A6--A7); draw(A1--A7--A2--A6); draw(A10--A4--A9--A5); dot("$A_1$",A1,dir(288)); dot("$A_2$",A2,dir(324)); dot("$A_3$",A3,dir(0)); dot("$A_4$",A4,dir(36)); dot("$A_5$",A5,dir(72)); dot("$A_6$",A6,dir(108)); dot("$A_7$",A7,dir(144)); dot("$A_8$",A8,dir(180)); dot("$A_9$",A9,dir(216)); dot("$A_{10}$",A10,dir(252)); [/asy][/asy]

The answer is $k\ge 5$. If $k\le 2$, for every $k$ of the points, there are less than $k$ total segments, so this is not possible. Note that there are a total of $45$ line segments. $~$ If $k=3$ then there is a color that appears at most $15$ times, $C$. Suppose for every set of three points, there is at least one edge colored $C$. Now, sum the number of edges $C$ for all sets of three points. Since each edge is counted eight times, we should receive at most $120$. Now, there are a total of $120$ sets of three points, so each set of three points must contain exactly one edge $C$ and there must be exactly $15$ edges $C$. Consider any edge $C$, then neither vertex at the endpoints of this edge can be connected to any other edge $C$. Therefore, only counting this one color, each vertex has degree at most one. However, that means at most $5$ appearances of this color, contradiction. Thus, there is a set of three points that does not contain this color. It is not possible. $~$ If $k=4$ then there is a color that appears at most $11$ times, $C$. Suppose for every set of four points, there is at least one edge colored $C$. We already know that there is a set of three points that no edge is colored $C$. Now, for each of the other seven points, the three edges between that point and the three selected points must have one colored $C$. Thus, among the seven, there are at most $4$ edges $C$. We once again pick a triangle such that there is no $C$ edge among them. For each of the four other points, there must be an edge connecting it and one of the points of the triangle that is $C$, leaving no more $C$ edges among the four remaining points, contradiction. $~$ If $k \ge 5$, for each color we will create $k-1$ groups of vertices, and we will color all the edges between vertices of the same group. In such a way, when we have $k$ vertices, for each color at least two vertices will be in the same group and thus the edge between them will be colored. For $k=5$ we color \begin{align*} C_1 &: 012 369 48 57 \\ C_2 &: 234 581 60 79 \\ C_3 &: 456 703 82 91 \\ C_4 &: 678 925 04 13 \\ C_5 &: 890 147 26 35 \end{align*}Note that not every edge is colored, but that's okay because we just color them however. For $k\ge 6$ the groups are easy: as long as the edges we color the same color never share an edge, we are fine.