MS_Kekas wrote: Positive real numbers $a_1, a_2, \ldots, a_{2024}$ are arranged in a circle. It turned out that for any $i = 1, 2, \ldots, 2024$, the following condition holds: $a_ia_{i+1} < a_{i+2}$. (Here we assume that $a_{2025} = a_1$ and $a_{2026} = a_2$). What largest number of positive integers could there be among these numbers $a_1, a_2, \ldots, a_{2024}$? If two successive $a_i$ are $\ge 1$, WLOG $a_1,a_2$ Then $a_3>a_1a_2\ge a_2\ge 1$ and sequence is stricly increasing from $a_2$, impossible since circular. So at most $1012$ positive integers. But in case of $1012$, sequence is $n_1,a_1,n_2,a_2,...,n_{1012},a_{1012}$ (where $n_i$ are positive integers) and we have $a_2>a_1n_1\ge a_1$ and sequence $a_i$ is stricly increasing, impossible since circular. So $\boxed{\text{At most }1011\text{ positive integers}}$ And indeed $1011$ positive integers is possible : $1,\frac 1{1012^2},1,\frac 2{1012^2},1,\frac 3{1012^2},...,1,\frac{1011}{1012^2},\frac 1{1012},\frac{1011.5}{1012^3}$