Romanian Master of Mathematics

The answer is in the affirmative. Note that the condition on $S$ may be rephrased as follows: For any subset $M$ of $N$ of size $|M|\ge 2$, there are at most $|M|-2$ sets in $S\cap P_3(M)$, where $P_3(M)$ is the set of 3-element subsets of $M$.

Extend the problem to $n\ge 1$ and proceed by induction on $n$; for formal correctness, assume that $|S|\le \max (0,n-2)$. The cases $n=1,2$ are both trivial.

Look upon points in the plane as complex numbers. Begin by considering a collection of points $q_1,q_2,\dots ,q_n$, not necessarily distinct, yet not all identical. Fix $q_n=0$ and associate with each $\{i,j,k\}$ in $S$ an equation $q_j-q_i=\omega \cdot (q_k-q_i)$, where $\omega =e^{i2\pi /3}$. This equation ensures that $q_i,q_j,q_k$ either are the vertices of a clockwise-oriented non-degenerate equilateral triangle or they all coincide.

We then get a system of $n-2$ linear equations with $n-1$ complex variables $q_1,q_2,\dots ,q_{n-1}$. It is well-known that such a system has a non-trivial solution $(q_1,q_2,\dots ,q_{n-1},q_n=0)$. However, some of the $q_i$ may coincide. So $N$ splits into some classes, $N=N_1\cup \cdots \cup N_m$, such that the points $q_i$ and $q_j$ coincide if and only if $i$ and $j$ share the same class and the size of each class is less than $n$. Note that the elements of any triple in $S$ either all lie in the same class or no two lie in the same class.

Clearly, each $S_i=S\cap P_3(N_i)$ satisfies the conditions in the statement relative to $N_i$ of size less than $n$. By the inductive hypothesis, for each $i=1,2,\dots ,m$, there exists a point-configuration $\{r_{ij}:j\in N_i\}$ satisfying the corresponding requirements. For each $j=1,2,\dots ,n$, choose the index $i$ of the class $N_i$ containing $j$, and let $p_j=q_j+ϵr_{ij}$, where $ϵ$ is small enough.

If the elements of a triple $\{i,j,k\}$ in $S$ lie in three different classes, then the angles of triangle $p_i{p}_j{p}_k$ are close to those of triangle $q_i{q}_j{q}_k$, provided that $ϵ$ is small enough,

so the triangle satisfies the required angle-condition. Otherwise, the three elements all lie in some class $N_{\ell }$, and the angles of $p_i{p}_j{p}_k$ equal those of $r_{\ell i}{r}_{\ell j}{r}_{\ell k}$, so they satisfy the requirements by the inductive hypothesis.

Remark. This solution proceeds via solving a linear system over ${\mathbb{F}}_3$ rather than $\mathbb{C}$. For each $\{i,j,k\}\in S$, we consider the equation $x_i+x_j+x_k=0$. Thus, we obtain $n-2$ equations in $n$ variables, so the solution space has dimension at least 2. It follows that there exists a solution for which $x_1,\dots ,x_n$ are not all equal. Let us define the sets $A_u=\{i\in [n]:x_i=u\}$ for $u\in {\mathbb{F}}_3$ corresponding to this solution. We then consider an equilateral triangle $P_0{P}_1{P}_2$ in the plane and place the points $p_i$ for $i\in A_u$ at $P_u$ for each $u\in {\mathbb{F}}_3$. As the $x_i$ solve the system of equations, all $\{i,j,k\}\in S$ satisfy $\{p_i,p_j,p_k\}=\{P_0,P_1,P_2\}$ or $p_i=p_j=p_k$. We may then proceed by induction by perturbing all of the points as in the first solution.