Question: Let $f(x)=\sin (\omega x+r \pi)$, where the constant $r \in$ Q. If $\omega \in \mathbf{Q}$ such that for any $n \in \mathbf{Z}$, $f(x)$ attains at least one maximum and one minimum value in the interval $[n, n+1]$, then what condition should $\omega$ satisfy?
Lemma Let $a \in \mathbf{R}, a \neq 0$, for any $M \geqslant 1$, there exists $m_{0} \in \mathbf{N}_{+}, m_{0} \leqslant[M], n_{0} \in \mathbf{Z}$, such that
$$
\left|m_{0} a-n_{0}\right|0$, let $m_{0}=j-i, n_{0}=[j a]-[i a]$;
if $j-i1$, let $T=1+d$, then $d>0$.

Without loss of generality, assume $\omega>0$.
Notice that, for $f(x)=1$, $x$ should satisfy
$$
\omega x+r \pi=2 m \pi+\frac{\pi}{2}(m \in Z) \text {. }
$$

Thus, the $x$ that makes $f(x)$ reach its maximum value is
$$
x_{m}=m \frac{2 \pi}{\omega}+\frac{\pi}{2 \omega}-\frac{r \pi}{\omega}=\left(\frac{2 m}{\omega}+\frac{1}{2 \omega}-\frac{r}{\omega}\right) \pi \notin \mathbf{Q}
$$
(at most one $x_{m^{\prime}}$ is an exception: $2 m^{\prime}+\frac{1}{2}-r=0$ ).
To prove there exist $m 、 n \in \mathbf{Z}$, such that
$$
\left\{\begin{array}{l}
x_{m}+\frac{T}{2}>n+1, \\
x_{m}-\frac{T}{2}\frac{1}{2}-\frac{T}{2}=-\frac{d}{2}, \\
x_{m}-\left(n+\frac{1}{2}\right)<\frac{T}{2}-\frac{1}{2}=\frac{d}{2}
\end{array}\right. \\
\Leftrightarrow\left|x_{m}-\left(n+\frac{1}{2}\right)\right|<\frac{d}{2} \\
\Leftrightarrow\left|\frac{2 m \pi}{\omega}-n+\frac{\pi}{2 \omega}-\frac{r \pi}{\omega}-\frac{1}{2}\right|<\frac{d}{2} \text {. } \\
\text { Let } \frac{\pi}{2 \omega}-\frac{r \pi}{\omega}-\frac{1}{2}=\left(\frac{1}{2 \omega}-\frac{r}{\omega}\right) \pi-\frac{1}{2}=\theta \text {, } \\
\end{array}
$$

then $\theta \neq 0$.
Thus, (2) $\Leftrightarrow\left|\frac{2 m \pi}{\omega}-n+\theta\right|<\frac{d}{2}$.
By the lemma, since $\frac{2 \pi}{\omega} \notin \mathbf{Q}$, there exist $m_{0} 、 n_{0} \in \mathbf{Z}$, such that
$$
0<\left|\frac{2 m_{0} \pi}{\omega}-n_{0}\right|<\frac{d}{2} \text {. }
$$

Therefore, there exists $k \in \mathbf{Z}$, such that
$$
\left|k\left(\frac{2 m_{0} \pi}{\omega}-n_{0}\right)+\theta\right|<\frac{d}{2}
$$
(use a "ruler" of length $\left|\frac{2 m_{0} \pi}{\omega}-n_{0}\right|$ to measure $\theta$ ).
Let $m=k m_{0}, n=k n_{0}$, then (1) holds.
At this point, $f(x)$ has only one maximum point $x_{m}$ in $[n, n+1]$, but no minimum point.
Thus, the required condition is $T=\frac{2 \pi}{|\omega|} \leqslant 1$, i.e., $|\omega| \geqslant 2 \pi$.