Example 3 Consider the permutation $\left(a_{1} a_{2} \cdots a_{20}\right)$ of $1,2 \cdots, 20$, and perform the following operation on this permutation: swap the positions of any two numbers, with the goal of transforming this permutation into $(1,2, \cdots, 20)$. Let the minimum number of operations required to achieve the goal for each permutation $a=\left(a_{1}, a_{2}, \cdots, a_{20}\right)$ be denoted as $k_{a}$. Find the maximum value of $k_{a}$.
Solution: For any permutation $a=\left(a_{1}, a_{2}, \cdots, a_{20}\right)$, it is clear that it can be transformed into $(1,2, \cdots, 20)$ in 19 operations, so $k_{a} \leqslant 19$.
Next, we prove that when $a=(2,3, \cdots, 20,1)$, at least 19 operations are required to transform it into $(1,2, \cdots, 20)$. For this, we introduce the concept of a cycle. Suppose in the permutation, the number at the $b_{1}$-th position is $b_{2}$, the number at the $b_{2}$-th position is $b_{3}, \cdots$, the number at the $b_{k-1}$-th position is $b_{k}$, and the number at the $b_{k}$-th position is $b_{1}$, then $b_{1} \rightarrow b_{2} \rightarrow \cdots \rightarrow b_{k} \rightarrow b_{1}$ is called a cycle. In particular, if the number at the $b_{1}$-th position is $b_{1}$, then $b_{1} \rightarrow b_{1}$ is also a cycle. Thus, each permutation can be decomposed into several non-intersecting cycles. Let the number of cycles in the permutation $a=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be denoted as $f(a)$. We examine the changes in $f(a)$ after each operation.
(1) When the operation is to swap two numbers $b_{i}, b_{j}(i<j)$ within the same cycle, then this cycle $b_{1} \rightarrow b_{2} \rightarrow \cdots \rightarrow b_{k-1} \rightarrow b_{k}$ is split into two cycles: $b_{1} \rightarrow b_{2} \rightarrow \cdots \rightarrow b_{i} \rightarrow b_{j+1} \rightarrow b_{j+2} \rightarrow \cdots \rightarrow b_{k} \rightarrow b_{1}$ and $b_{i+1} \rightarrow b_{i+2} \rightarrow \cdots \rightarrow b_{j} \rightarrow b_{i+1}$.
(2) When the operation swaps two numbers $b_{1}$ and $c_{1}$ from two cycles $b_{1} \rightarrow b_{2} \rightarrow \cdots \rightarrow b_{k} \rightarrow b_{1}$ and $c_{1} \rightarrow c_{2} \rightarrow \cdots \rightarrow c_{m} \rightarrow c_{1}$, these two cycles merge into one cycle: $b_{1} \rightarrow c_{2} \rightarrow c_{3} \rightarrow c_{m} \rightarrow c_{1} \rightarrow b_{2} \rightarrow b_{3} \rightarrow \cdots \rightarrow b_{k} \rightarrow b_{1}$.

Therefore, after each operation, the number of cycles always increases by 1 or decreases by 1. Note that $a=(2,3, \cdots, 20,1)$ has only one cycle $1 \rightarrow 2 \rightarrow \cdots \rightarrow 20 \rightarrow 1$, while $(1,2, \cdots, 20)$ has 20 cycles, $1 \rightarrow 1,2 \rightarrow 2, \cdots, 20 \rightarrow 20$. Therefore, transforming $a$ into $(1,2, \cdots, 20)$ requires at least 19 operations.
Thus, the maximum value of $k_{a}$ is 19.