Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}<a_{2}<\cdots<a_{100}$) and the other in descending order (denoted as $b_{1}>b_{2}>\cdots>b_{100}$). Try to find
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{100}-b_{100}\right|
$$

the value of the expression.
First, prove: For any term in the algebraic expression
$$
\left|a_{k}-b_{k}\right|(k=1,2, \cdots, 100)
$$

the larger number among $a_{k}$ and $b_{k}$ must be greater than 100, and the smaller number must not exceed 100.
(1) If $a_{k} \leqslant 100$ and $b_{k} \leqslant 100$, then by
$$
a_{1}b_{k+1}>\cdots>b_{100}
$$
we get that $a_{1}, a_{2}, \cdots, a_{k}, b_{k}, b_{k+1}, \cdots, b_{100}$ are 101 numbers, all of which do not exceed 100, which is impossible;
(2) If $a_{k}>100$ and $b_{k}>100$, then by
$$
a_{100}>a_{99}>\cdots>a_{k+1}>a_{k}>100
$$

and $b_{1}>b_{2}>\cdots>b_{k}>100$
we get that $b_{1}, b_{2}, \cdots, b_{k}, a_{k}, a_{k+1}, \cdots, a_{100}$ are 101 numbers, all of which are greater than 100, which is also impossible.
Thus, in the 100 absolute values of the algebraic expression
$$
\left|a_{1}-b_{1}\right|,\left|a_{2}-b_{2}\right|, \cdots,\left|a_{100}-b_{100}\right|
$$

the smaller numbers are $1,2, \cdots, 100$, and the larger numbers are 101, $102, \cdots, 200$.
Therefore, the original expression is
$$
\begin{array}{l}
=(101+102+\cdots+200)-(1+2+\cdots+100) \\
=10000 .
\end{array}
$$