Example 1 Given $2 n$ real numbers: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$. Let
$$F=a_{1} b_{i_{1}}+a_{2} b_{i_{2}}+\cdots+a_{n} b_{i_{n}}$$

where $b_{i_{1}}, b_{i_{2}}, \cdots, b_{i_{n}}$ is a permutation of $b_{1}, b_{2}, \cdots, b_{n}$. Find the maximum and minimum values of $F$
If the expression of $F$ does not contain the term $a_{1} b_{1}$, then consider the two terms containing $a_{1}$ and $b_{1}$ respectively: $a_{1} b_{i_{1}}$ and $a_{j} b_{1}$. Adjust them to $a_{1} b_{1}$ and $a_{j} b_{i_{1}}$, respectively. Then,
$a_{1} b_{1} + a_{j} b_{i_{1}} - (a_{1} b_{i_{1}} + a_{j} b_{1}) = (a_{1} - a_{j})(b_{1} - b_{i_{1}}) \geqslant 0$ (polishing tool).
It can be seen that after adjusting to include the term $a_{1} b_{1}$, the value of $F$ does not decrease.
Fix this product $a_{1} b_{1}$, and then consider the remaining $n-1$ products. In this way, after at most $n-1$ adjustments, the terms $a_{1} b_{1}, a_{2} b_{2}, \cdots, a_{n} b_{n}$ will appear, and the value of $F$ remains non-decreasing. Thus, $F_{1} = a_{1} b_{1} + a_{2} b_{2} + \cdots + a_{n} b_{n}$ is the maximum value of $F$.

Similarly, when the terms $a_{1} b_{n}, a_{2} b_{n-1}, \cdots, a_{n} b_{1}$ appear, $F$ reaches its minimum value.