Given,
$$
\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}=\frac{\sqrt{n}}{2}
$$

Squaring both sides
$$
\left(\sin \frac{\pi}{2 n}+\cos \frac{\pi}{2 n}\right)^2=\frac{n}{4}
$$


$$
\begin{gathered}
\sin ^2 \frac{\pi}{2 n}+\cos ^2 \frac{\pi}{2 n}+2 \sin \frac{\pi}{2 n} \cos \frac{\pi}{2 n}=\frac{n}{4} \\
\Rightarrow 1+\sin \frac{\pi}{n}=\frac{n}{4} \Rightarrow \sin \frac{\pi}{n}=\frac{n}{4}-1 \Rightarrow \sin \frac{\pi}{n}=\frac{n-4}{4}
\end{gathered}
$$

As, $\quad \sin \frac{\pi}{n}>0, n>2$
So, $\quad \frac{n-4}{4}>0 \Rightarrow n>4$
Also, $\sin \left(\frac{\pi}{2 n}\right)+\cos \frac{\pi}{2 n}=\sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)$.
From Eq. (i)
$\sqrt{2} \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2} \quad \sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right)=\frac{\sqrt{n}}{2 \sqrt{2}}$
Since, $\sin \left(\frac{\pi}{4}+\frac{\pi}{2 n}\right) < 1, \forall n>2$
We get, $\frac{\sqrt{n}}{2 \sqrt{2}} < 1 \Rightarrow n < 8$
From Eqs. (ii) and (iii)
$$
4 < n < 8
$$

So, $n=5,6,7$
Hence, number of integral value of $n$ is 3 .