Example 12 Let $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a non-decreasing sequence of positive integers. For $m \geqslant 1$, define $b_{m}=\min \left\{n, a_{n} \geqslant m\right\}$, i.e., $b_{m}$ is the smallest value of $n$ such that $a_{n} \geqslant m$. Given that $a_{19}=85$, find
$$a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+b_{85}$$

the maximum value.
If there exists an $i$ such that $a_{i}<a_{i+1}(1 \leqslant i \leqslant 18)$, then perform the following adjustment: $a_{i}^{\prime}=a_{i}+1$, $a_{j}^{\prime}=a_{j}(j \neq i)$, and denote the adjusted $b_{j}$ as $b_{j}^{\prime}(j=1,2, \cdots, 85)$.

By definition, $b_{a_{i}+1}=i+1, b_{a_{i}+1}^{\prime}=i=b_{a_{i}+1}-1, b_{j}^{\prime}=b_{j}\left(j \neq a_{i}+1\right)$.
Therefore, the above adjustment decreases $b_{a_{i}+1}$ by 1 while keeping the other $b_{j}$ unchanged, hence this adjustment maintains the value of (1) unchanged.

Thus, a series of adjustments can be made to achieve $a_{1}=a_{2}=\cdots=a_{19}=85$ while keeping the value of (1) unchanged. At this point, $b_{1}=b_{2}=\cdots=b_{85}=1$, so the maximum value of the expression (1) is
$$19 \times 85 + 1 \times 85 = 20 \times 85 = 1700 \text{.}$$