For every positive integer $x$, obtain the number $z$ according to the following process:
(i) Move the last digit of $x$ to the first position to get $y$;
(ii) Then take the square root of $y$ to get its arithmetic square root $z$.
For example, if $x=9, y=9, z=\sqrt{9}=3$;

and if $x=5002, y=2500$,
$$
z=\sqrt{2500}=50 \text {. }
$$

Find all positive integers $x$ such that $z=\frac{1}{2} x$.
(1) If $x$ is a one-digit number, then $y=x, z=\sqrt{y}=\sqrt{x}$.

Also, $z=\frac{1}{2} x$, so $\sqrt{x}=\frac{1}{2} x$.
Solving this, we get $x_{1}=4, x_{2}=0$ (discard).
Thus, $x=4$.
(2) If $x$ is a two-digit number, let $x=10 a+b, a, b$ be integers, and $0<a \leqslant 9,0 \leqslant b \leqslant 9$. Then
$$
y=10 b+a, z=\sqrt{10 b+a} .
$$

Also, $z=\frac{1}{2} x$, so,
$$
\sqrt{10 b+a}=\frac{1}{2}(10 a+b) \text {. }
$$

Since $a, b$ are integers, $\frac{1}{2}(10 a+b)$ is a rational number, thus, $10 b+a$ must be a perfect square, and $\sqrt{10 b+a} \leqslant 9$. Therefore, $\frac{1}{2}(10 a+b)$ is also an integer no greater than 9. Hence, $a=1$. Thus, we have
$$
\sqrt{10 b+1}=5+\frac{1}{2} b .
$$

Solving this, we get $b_{1}=8, b_{2}=12$ (discard).
Thus, $x=18$.
(3) If $x \geqslant 100$, then $y<10 x$,
$$
\begin{array}{l}
z=\sqrt{y}<\sqrt{10 x}=\frac{\sqrt{10}}{\sqrt{x}} \cdot x \\
\leqslant \frac{\sqrt{10}}{\sqrt{100}} x=\frac{1}{\sqrt{10}} x<\frac{1}{2} x .
\end{array}
$$

Thus, $x \geqslant 100$ does not meet the condition.
Therefore, the positive integer $x=4$ or 18.