It does hold that for any constant $c$ and sufficiently large $n>N(c)$, any coloring of more than $cn$ vertices of a regular $(4n+1)$-gon yields colored isosceles triangle. So I hope this problem has a considerably good and short solution for this specific choices of $n\geqslant 3$.

sqing wrote: Let $n$ be an integer with $n\geq 3$ and assume that $2n$ vertices of a regular $(4n + 1)-$gon are coloured. Show that there must exist three of the coloured vertices forming an isosceles triangle. Assume the contrary. Label the vertices $P_1, P_2, P_3, P_4, ..., P_{4n}, O$ in that order. Note that $P_i, O, P_{4n+1-i}$ from an isoceles triangle. So, we must have atleast two consecutive vertices to be coloured. WLOG, let the vertices be $O, P_1$. It is easy to see that no two consecutive vertices in the sequences $$ P_{4n-1}, \: P_4,\: P_{4n-3},\: P_{6},\: P_{4n-5}, \: ...,\: P_{2n-2},\: P_{2n+3},\: P_{2n}$$and $$P_{3},\: P_{4n-2},\: P_5,\: P_{4n-4},\: P_7,\: ...,\: P_{2n+2},\: P_{2n-1}$$are both coloured. So, WLOG assume that only the odd terms of both sequences are coloured. But this means that $O, P_{4n-1}, P_{4n-1}$ are coloured. A contradiction. So, there exists an “colourful” isoceles triangle. $\blacksquare$

Assume Contrary Let $P_{-2n},P_{-2n+1},...,P_0,P_1,...,P_{2n-1},P_{2n}$ be vertices of $(4n+1)$ -gon. 1.case There are two coloured neighboring vertices. WLOG:$P_0,P_1$ are coloured. Then at most one of $P_i$ and $P_{-i}$ can be coloured $(i=1,2,...,2n)$ Since $P_0P_iP_{-i}$ form isosceles triangle And at most one of $P_{i+2}$ and $P_{-i}$ can be coloured $(i=1,2,...,2n-2)$ Since $P_1P_{i+2}P_{-i}$ form isosceles triangle Also note that 2 fact : at most one of $P_{-2n}$ and $P_{-2n+1}$ can be colored. $P_{-1},P_2,P_{-2n}$ are not coloured since they form isosceles triangle with $P_0$ and $P_1$ So no consecutive vertices in two sequences : $$P_{-2} - P_{4} - P_{-4} - ... - P_{-2n+2} - P_{2n}$$$$P_{3} - P_{-3} - P_{5} - ... - P_{2n-1} - P_{-2n+1}$$are coloured. At most half of both sequence can be coloured.After counting we see at most $2n-2$ vertices can be coloured in sequences(together). Since we have $2n$ coloured vertices and two of them are $P_0,P_1$ we get that excatly half of vertices are coloured in both sequence. Now we are going to look 3 triangles : $P_0P_{-2}P_{-4} \implies P_{-2},P_{-4}$ not coloured $\implies P_{2n}$ coloured $\implies P_{2n-2}$ coloured$(1)$ $P_1P_{3}P_{5} \implies P_{3},P_{5}$ not coloured $\implies P_{-2n+1} $ coloured $P_{2n-2}P_{2n}P_{-2n+1} \implies P_{2n-2} $ not coloured$(2)$ By 1 and 2 we get Contradiction or Case1 2.case There are not any two coloured neighboring vertices WLOG: $P_0$ is coloured Then $P_1,P_{-1},P_{2n},P_{-2n}$ are not coloured.So easily we get at most $2n-1$ vertices can be colored Thus we get Contradiction for Case2 So we are done!