Example 3 Arrange the numbers $1,2,3, \cdots, 64$ in any order as $a_{1}, a_{2}, a_{3}, \cdots a_{64}$ (1), calculate $\left|a_{1}-a_{2}\right|,\left|a_{3}-a_{4}\right|, \cdots,\left|a_{63}-a_{64}\right|$; then rearrange these 32 numbers in any order as $b_{1}, b_{2}, b_{3}, \cdots b_{32}$ (2), calculate $\left|b_{1}-b_{2}\right|,\left|b_{3}-b_{4}\right|, \cdots,\left|b_{31}-b_{32}\right|$, $\cdots$, and continue this process until you get a single number $x$. Try to determine the parity of $x$.
Solution: We know that for $a \in Z$, we have $a \equiv -a(\bmod 2),|a| \equiv a(\bmod 2)$. Using the above conclusions, consider the sum of the numbers in (1) and the sum of the numbers in (2),
$$
\begin{array}{l}
b_{1}+b_{2}+b_{3}+\cdots+b_{32} \\
=\left|a_{1}-a_{2}\right|+\left|a_{3}-a_{4}\right|+\cdots+\left|a_{63}-a_{64}\right| \\
=a_{1}-a_{2}+a_{3}-a_{4}+\cdots+a_{63}-a_{64} \\
\equiv a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{63}+a_{64}(\bmod 2)
\end{array}
$$

The above equation shows that although after one "operation", (1) becomes (2), the parity of the sum remains unchanged. Of course, this is also true after multiple operations, so
$$
x \equiv a_{1}+a_{2}+a_{3}+\cdots+a_{63}=1+2+3+\cdots+64 \equiv 0(\bmod 2)
$$

That is, $x$ is even.