Problem 2317

Consider the 2002 intervals $I_k = (\frac{k}{2002}, \frac{k+1}{2002})$ where $k = -1001, -1000, \dots, 1000$. Since $y$ is irrational, the numbers $y - y'$, $2y - (2y)'$, $\dots$, $2001y - (2001y)'$ are irrational numbers between $-\frac{1}{2}$ and $\frac{1}{2}$. Thus, each of them belongs to one of the $I_k$'s.

We claim that one of the numbers $my - (my)'$ (with $1 \le m \le 2001$) belongs to $I_{-1}$ or $I_0$. Suppose on the contrary that all these numbers belong to the other 2000 intervals. By the pigeonhole principle, two of them, say $my - (my)'$ and $ny - (ny)'$ with $m > n$, belong to the same interval $I_k$. Then we have
$$(m-n)y - [(my)' - (ny)'] = [my - (my)'] - [ny - (ny)'] < \frac{k+1}{2002} - \frac{k}{2002} = \frac{1}{2002}.$$
Therefore, $(my)' - (ny)'$ is the closest integer to $(m-n)y$, and we know that $(m-n)y - [(m-n)y]'$ belongs to $I_{-1}$ or $I_0$. This is a contradiction. Therefore,

we can find some $my - (my)'$ belonging to $I_{-1}$ or $I_0$. For this $m$, we have
$$\langle my \rangle = |my - (my)'| < \frac{1}{2002} < \frac{1}{2001}.$$
This completes the proof.