Example 10 Let $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a sequence of non-decreasing 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 $a_{n} \geqslant 3$. Given that $a_{19}=85$, find the maximum value of
$$
a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+b_{85}
$$
(1985 USA Olympiad Problem)
If there exists $i(1 \leqslant i \leqslant 18)$, such that $a_{i}<a_{i+1}$, then perform the following adjustment: $a_{i}{ }_{i}=a_{i}+1, a^{\prime}{ }_{j}=a_{j}$ $(j \neq i)$, and denote the adjusted $b_{j}$ as $b_{j}^{\prime}, j=1,2, \cdots, 85$. Thus, we can see that
$b_{a_{i+1}}=i+1, b_{a_{i+1}}^{\prime}=i=b_{a_{i+1}}-1, b_{j}^{\prime}=b_{j}, j \neq a_{i}+1$.
This means that the adjustment decreases $b_{a_{i+1}}$ by 1 while keeping the other $b_{j}$ unchanged, therefore, the adjustment maintains the value of $(*)$.

In this way, we can perform a series of adjustments to make $a_{1}=a_{2}=\cdots=a_{19}=85$ while keeping the value of $(*)$ unchanged, but at this point $b_{1}=b_{2}=\cdots=b_{85}=1$. Therefore, the maximum value of $(*)$ is
$$
19 \cdot 85+1 \cdot 85=20 \cdot 85=1700 \text {. }
$$