On when does the period of the discrete-time sinusoid decrease as the frequency increase.
For a discrete-time sinusoid $x[n]$ of frequency $f_0$
$$
x[n] = x[n + N] \iff f_0 = \frac kN\tag{1}
$$
where $k$ and $N$ are relatively prime integers. The condition in Equation $(1)$ is tantamount to
$$
\omega_0 N = 2\pi k \tag{2}
$$
What to note down is that discrete-time signals are defined only for integer indices $n$. And that $k$ in Equation $(1)$ and $(2)$ correspond to an integer number of periods (i.e. $kT_s$) of the corresponding continuous-time sinusoid.
With that in mind; if the frequency of a discrete-time sinusoid is varied from $\omega_1$ with period $N_1$ to $\omega_2$ with period $N_2$, and they both fit in exactly one period of their corresponding continuous-time equivalents, meaning
$$
\frac{2\pi}{\omega_1}\in \mathbb N\quad\text{and}\quad\frac{2\pi}{\omega_2}\in \mathbb N\tag{3}
$$
then it can be seen that
$$
\omega_2 > \omega_1 \implies N_2 < N_1\tag{4}
$$
A quick check can be made using the values in the table below. The values of $2\pi/\omega_0$ for which the discrete-time sinusoid fit in exactly one period of their corresponding continuous-time equivalent are highlighted in blue.
$$
\begin{array}{cccc}
\hline
\omega_0 & f_0 & N & \frac{2\pi}{\omega_0}\\ \hline
0 & \infty & 0 & \infty\\
\frac{3\pi}{32} & \frac{3}{64} & 64 & \frac{64}{3}\\
\frac{\pi}{8} & \frac{1}{16} & 16 & \color{blue}{16}\\
\frac{\pi}{4} & \frac{1}{8} & 8 & \color{blue}{8}\\
\frac{3\pi}{8} & \frac{3}{16} & 16 & \frac{16}{3}\\
\frac{\pi}{2} & \frac{1}{4} & 4 & \color{blue}{4}\\
\frac{29\pi}{30} & \frac{29}{60} & 60 & \frac{60}{29}\\
\frac{999\pi}{1000} & \frac{999}{2000} & 2000 & \frac{2000}{999}\\
\pi & \frac{1}{2} & 2 & \color{blue}{2}\\\hline
\end{array}
$$
In all other cases, increasing the value of $\omega_0$ does not necessarily decrease the period of the signal.