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Claim. For every $N\ge4$, with $n$ the integer such that $2^{n-1}<N\le2^n$,

$$\alpha_N=\pi\Bigl(1-\frac1n\Bigr) \quad\text{for }2^{n-1}+2^{n-3}<N\le2^n,$$ $$\alpha_N=\pi\Bigl(1-\frac2{2n-1}\Bigr) \quad\text{for }2^{n-1}<N\le2^{n-1}+2^{n-3}.$$

Here $\alpha_N$ is the supremum of the angles $\alpha$ such that every set of $N$ points in the plane has three distinct points determining an angle of at least $\alpha$, the quantity the problem asks to determine. The formula is taken from the zbMATH record of the 1993 note, reindexed from the record's convention $2^n<N\le2^{n+1}$ to the range $2^{n-1}<N\le2^n$ used here. The record prints the two branches with no lower limit on $N$, but the formula holds only from $N=4$ on: at $N=3$ its first branch would give $\pi/2$, while $\alpha_3=\pi/3$, the equilateral triangle, as Erdős and Szekeres record. From $N=4$ on it reproduces their values $\alpha_4=\pi/2$, $\alpha_5=3\pi/5$ and $\alpha_6=\alpha_7=\alpha_8=2\pi/3$. The site's commentary prints $1/(2n-1)$ in place of $2/(2n-1)$ on the second branch, which is inconsistent with $\alpha_5=3\pi/5\le\alpha_8=2\pi/3$ (with $1/(2n-1)$ it would give $\alpha_5=4\pi/5$).

History. The problem goes back to Blumenthal. Szekeres (1941) proved $\alpha_{2^n}\le\pi(1-1/n)$ and the lower bound $\alpha_{2^n+1}>\pi(1-1/n+1/(n(2^n+1)^2))$. Erdős and Szekeres (1960) proved $\alpha_{2^n}=\alpha_{2^n-1}=\pi(1-1/n)$ for $n\ge3$, recorded on their claim page (their print states the $2^n-1$ case for $n\ge2$, an error, since $\alpha_3=\pi/3$), and suggested that $\alpha_N=\pi(1-1/n)$ might hold throughout $2^{n-1}<N<2^n$ for $n\ge4$. Sendov refuted that suggestion in a first note, On a conjecture of P. Erdős and G. Szekeres, C. R. Acad. Bulgare Sci. 45 (1992), no. 12, 17–20, and then determined every value in the paper this page records. The values at powers of two agree with Erdős and Szekeres, and the second branch of the formula is where the suggestion fails. Sendov's full paper, Minimax of the angles in a plane configuration of points, Acta Math. Hungar. 69 (1995), no. 1–2, 27–46, proves the same theorem; its zbMATH record (Zbl 0853.52009) prints the formula with $n\ge2$ in the record's convention, that is from $N=5$ on, and takes $\alpha_3$, $\alpha_4$ and $\alpha_5$ as known.

Acceptance. The result is refereed twice: the 1993 note, Bl. Sendov, Angles in a plane configuration of points, C. R. Acad. Bulgare Sci. 46 (1993), no. 5, 27–30, cited as its zbMATH record gives it, and the full publication with proofs, Bl. Sendov, Minimax of the angles in a plane configuration of points, Acta Mathematica Hungarica 69 (1995), no. 1–2, 27–46, the paper link above. The curator of erdosproblems.com, Thomas Bloom, labels the problem solved and credits the 1993 note with the definitive answer. Neither Comptes rendus note is available online; the links above are the zbMATH record of the 1993 note, the 1995 paper's DOI and the site's page, and the page is dated to the 1993 note's publication year, the record giving no day. No check of the proof beyond the journals' refereeing and the curator's credit is recorded.