$2 \cdot 96$ Let $a_{1}, a_{2}, a_{3}, \cdots$ be a non-decreasing sequence of positive integers. For $m \geqslant 1$, define
$$
b_{m}=\min \left\{n \mid a_{n} \geqslant m\right\},
$$

i.e., $b_{m}$ is the smallest value of $n$ such that $a_{n} \geqslant m$.
If $a_{19}=85$, find the maximum value of $a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+b_{85}$.
［Solution］In general, let $a_{q}=p$, we prove
$$
a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2}+\cdots+b_{p}=p(q+1)
$$

For this problem, when $q=19, p=85$,
$$
p(q+1)=85(19+1)=1700 .
$$
(i) If $a_{1}-a_{2}=\cdots=a_{q}=p$, then by definition
$$
b_{1}=b_{2}=\cdots=b_{p}=1
$$

Thus, we have
$$
a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2}+\cdots+b_{p}=p q+p=p(q+1) .
$$
(ii) If $a_{i}<p(1 \leqslant i<q)$.

Let $t$ be the largest index such that $a_{t}<p$, and let $a_{t}=u$.
If $a_{t}$ is increased by 1, then $b_{j}(j \neq u+1)$ remains unchanged, while $b_{u+1}$ decreases by 1, so the total sum in the problem remains unchanged.
We repeat this process until
$$
a_{1}=a_{2}=\cdots=a_{q}=p .
$$

By (1), we get
$$
a_{1}+a_{2}+\cdots+a_{q}+b_{1}+b_{2}+\cdots+b_{p}=p(q+1) .
$$