let we use the usual setup let $n = m^2+k , 0\leq k \leq 2m $ $m-1 \mid m^2+k$ $m-1 \mid k+1$ $m =1 $, no sols $m = 2,k=0,1,2,3,4$ work $m=3 , k = 1,3,5$ $m=4 , k = 2,5,8$ for $m \geq 5$ $a(m-1) = k+1 \leq 2m+1$ where a is a positive integer $a \leq \frac{2m+1}{m-1} \leq 2$ so $2m-3 = k$ or $m-2 = k$ for each $m \geq 5$ $2023 \geq m^2$ , $m \leq 44$ so from 5-44 , there is 40 number , 40*2 = 80 +5+3+3 so the total num is 91

Tintarn wrote: How many positive integers $n<2024$ are divisible by $\lfloor \sqrt{n}\rfloor-1$? Suppose $n\in [a^2,(a+1)^2), a\in \mathbb{N}$. Hence $\lfloor n \rfloor-1=a-1$. From $(\lfloor \sqrt{n}\rfloor-1) | n$ we have $\frac{n}{a-1}\in \mathbb{Z}$, which leads to $$\frac{n}{a-1}\ge \frac{a^2}{a-1}=a+1+\frac{1}{a-1}, \frac{n}{a-1}\le\frac{a^2+2a+1}{a-1}=a+3+\frac{4}{a-1}\Leftrightarrow a+2\le \frac{n}{a-1}\le a+3$$. So when $a\ge 6$, $n=a^2+a-2 \;\text{or} \;a^2+2a-3$. When $a\ge 6$, there are 2 $n$s in $[a^2,a^2+2a+1)$; when $a\le 5$, $n=4,5,6,7,8,10,12,14,18,21,24,28,32$ can satisfy the statement. In that case, there are $\boxed{91}\; n$s in total. $\square$ [EDIT: wow i'm just a few seconds slower than @above what luck]

StarLex1 wrote: let we use the usual setup let $n = m^2+k , 0\leq k \leq 2m $ $m-1 \mid m^2+k$ $m-1 \mid k+1$ $m =1 $, no sols $m = 2,k=1,2,3,4$ work $m=3 , k = 1,3,5$ $m=4 , k = 2,5,8$ for $m \geq 5$ $a(m-1) = k+1 \leq 2m+1$ where a is a positive integer $a \leq \frac{2m+1}{m-1} \leq 2$ so $2m-3 = k$ or $m-2 = k$ for each $m \geq 5$ $2023 \geq m^2$ , $m \leq 44$ so from 5-44 , there is 40 number , 40*2 = 80 +4+3+3 so the total num is 90 m=2,k=0 should work(n=4)

REDACTED, i accidentally clicked the submit im so sorry

oh yeah i miscounted , i start counting from 1 , lol