2 In a non-decreasing sequence of positive integers $a_{1}, a_{2}, \cdots, a_{m}, \cdots$, for any positive integer $m$, define $b_{m}=$ $\min \left\{n \mid a_{n} \geqslant m\right\}$. It is known that $a_{19}=85$. Find the maximum value of $S=a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+$ $b_{85}$. (1985 USA Mathematical Olympiad Problem)
2. First, the largest number must exist. From the condition, we have $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{19}=85$. We conjecture that the extremum point is when each $a_{i}$ is as large as possible and all $a_{i}$ are equal, and all $b_{j}$ are equal. In fact, if there is $a_{i}a_{i}$, so $a_{i+1} \geqslant a_{i}+1$. But $a_{i}<a_{i}+1$, so $b_{a_{i}+1}=i+1$, $b_{a_{i}+1}^{\prime}=i=b_{a_{i}+1}-1, b_{j}^{\prime}=b_{j}$ (when $j \neq a_{i}+1$). From this, we can see that the adjustment makes $b_{a_{i}+1}$ decrease by 1, and the rest of the $b_{i}$ remain unchanged. Thus, the value of $S$ does not decrease. In summary, we have $S \leqslant 19 \times 85 + 1 \times 85 = 1700$, with equality when $a_{i}=85, b_{j}=1(1 \leqslant i \leqslant 19,1 \leqslant j \leqslant 85)$.