let each vertice of the regular $2024$-gon be $a_1,a_2,..,a_{2024}$ define $f(a_n,a_m)$ as the number of vertice between $a_n$ and $a_n$ which $f(a_n,a_m)\leq 1011~~~,\forall n,m\in\{1,2,..,2024\}$ let $b_1,b_2,..,b_k\in\{1,2,..,2024\}$ which $b_1<b_2<b_3<..<b_k$ such that $a_{b_1}a_{b_2}...a_{b_k}$ form a regular $k$-gon we can clearly see that $a_{n}a_m=a_pa_q\Longleftrightarrow f(a_n,a_m)=f(a_p,a_q)$ thus $f(a_{b_1},a_{b_2})=f(a_{b_2},a_{b_3})=...=f(a_{b_{k-1}},a_{b_k})=f(a_{b_k},a_{b_1})~~~(*)$ $\Rightarrow k|f(a_{b_1},a_{b_2})+f(a_{b_2},a_{b_3})+...+f(a_{b_k},a_{b_1})$ since $f(a_{b_1},a_{b_2})+f(a_{b_2},a_{b_3})+...+f(a_{b_k},a_{b_1})+k=2024$ we'll get that $~k|2024$ hence, $k=4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024$ we can observe that if any regular $ k$-gon has a marked vertice for$k=4,11,23$ then the problem is true thus for any regular $k$-gon let the vertices be $a_{b_1},a_{b_2},..,a_{b_k}$ from $(*)$ we'll get that $(b_1,b_2,..,b_k)=(b_1,b_1+\frac{2024}{k},b_1+2(\frac{2024}{k}),..,b_1+(k-1)(\frac{2024}{k}))$ $\Rightarrow b_1\equiv b_2\equiv...\equiv b_k(mod~\frac{2024}{k})$ therefore for all $q$ in CRS of $\frac{2024}{k}$ there exist $p\in\{1,2,..,2024\}$ which $p\equiv q(mod~\frac{2024}{k})$ and vertice $a_p$ is marked for all $k=4,11,23$ thus atleast $\frac{2024}{4}=506$ vertices are marked $\therefore$ the minimal number of vertices marked is $506~$ which can be achieve by marking $a_{2024},a_1,a_2,..,a_{505}~~$Q.E.D$~\square$ lmk if there is any mistakes i still think my solution is still too long and would waste too much time tho

I am not sure what happened here, but clearly if you divide all vertices on 506 squares you get a bound for 506, so 505 can't be right

atdaotlohbh wrote: I am not sure what happened here, but clearly if you divide all vertices on 506 squares you get a bound for 506, so 505 can't be right thanks for letting me know all fixed the lack of sleep is really getting to me