Take
Fix some algebraic curve $\gamma$ if natural number $\mathbb R^4$ such that for any 4 distinct points N$.Choose a half-circle $x,y,v,w$ we have \gamma$. Let $y-x+w-v\ne 0$.Consider all M$ be the number of integer points on $\gamma$One can choose $\gamma$ so that there are $M$ is quite many big;say $M\gg C\cdot N^{2/5}$ is easy to arrange, but one can do better.
Note that for any two distinct pairs $(x,y)$ and $(v,w)$ of these integer points .on $\gamma$ we have$$|(y-x)+(w-v)|\ge 1.$$
For each integer point $x=(x_1,x_2,x_3,x_4)$ on $\gamma$ x=(x_1,x_2)\in\gamma$, consider the number$$\bar x=x_1+(2\cdot N)\cdot x_2+(2\cdot N)^2\cdot x_3+(2\cdot N)^3\cdot x_4$$All $\bar x$'s form a a set $\bar X$ of integers in $[0,(2\cdot N)^4]$ N)^2]$ such that $$\bar y-\bar x+\bar w-\bar v\ne 0.$$
Now choose small for any two distinct pairs $\varepsilon>0$ (\bar x,\bar y)$ and mark $(\bar v,\bar w)$ of numbers in $\bar X$, $$|(\bar y - \bar x)+(\bar w - \bar v)| \ge 1.$$
Mark all points on a circle an with central angles of the form $\tfrac{\bar x\cdot\pi}{(2\cdot N)^2}$, $\bar x\cdot\varepsilon$x\in \bar X$.You all Note that the angles between the lines through these points will be $\ge \varepsilon$.
For tfrac{\pi}{2\cdot (2\cdot N)^2}$.
If you look for a set of $n$ points, you it should give a bound on sufficient to take $N\gg n^{5/2}$ and therefore the angle of the form will be $O(1/n^{2+?})$;an obvios upper bound is \gg 1/n^5$.This estimate can be improved, but I do not think you can get $O(1/n^2)$.\gg 1/n^2$ on this way;this might be the optimal asymptotic (?).

