Let $\left\{a_{n}\right\}$ be a sequence of positive integers that is monotonically increasing, that is,

$$
1 \leqq a_{1} \leqq a_{2} \leqq \ldots \leqq a_{n} \leqq \ldots
$$

For any integer $m \geqq 1$, let $b_{m}$ be the index of the first element in the sequence $\left\{a_{n}\right\}$ that is at least as large as $m$.

If $a_{19}=88$, then between what bounds can the sum

$$
a_{1}+a_{2}+\ldots+a_{19}+b_{1}+b_{2}+\ldots+b_{88}
$$

vary?
I. Solution. We will show that the sum in question is the same for every sequence $\left\{a_{n}\right\}$. First, note that since $a_{19}=88$, the values $b_{1}, b_{2}, \ldots, b_{88}$ exist and each is at most 19.

![](https://cdn.mathpix.com/cropped/2024_05_02_56d4735d9709920ac5cdg-1.jpg?height=391&width=636&top_left_y=273&top_left_x=734)

Construct the $19 \times 88$ table shown in the figure. Color the cell in the $i$-th row and $j$-th column black if $j \leq a_{i}$, otherwise leave the cell white.

In the $i$-th row, there are clearly $a_{i}$ black cells, so in the table, we have colored

$$
a_{1}+a_{2}+\ldots+a_{19}
$$

cells black.

Count the white cells column by column. In the $j$-th column, the cells in the rows where $a_{i} < j$ remain white; the number of these is exactly the number of elements in the sequence $\left\{a_{n}\right\}$ that are less than $j$, which, by the definition of $\left\{b_{n}\right\}$, is precisely $b_{j}-1$. Therefore, the total number of white cells is

$$
\left(b_{1}-1\right)+\left(b_{2}-1\right)+\ldots+\left(b_{88}-1\right)
$$

Every cell in the $19 \times 88$ table is either black or white, so the value of the sum in question is

$$
19 \cdot 88 + 88 = 1760
$$

for every monotonically increasing sequence $\left\{a_{n}\right\}$ such that $a_{19}=88$.

II. Solution. By the definition of the sequence $\left\{b_{n}\right\}$, the sequence $\left\{a_{n}\right\}$ has $b_{m}-1$ elements less than $m$, so $\left(b_{m+1}-1\right)-\left(b_{m}-1\right)=b_{m+1}-b_{m}$ elements are exactly $m$.

Since $a_{19}=88$ by the condition, the number 88 clearly appears $\left(20-b_{88}\right)$ times. This means that

$$
\begin{gathered}
a_{1}+a_{2}+\ldots+a_{19}= \\
=\left(b_{2}-b_{1}\right) \cdot 1+\left(b_{3}-b_{2}\right) \cdot 2+\ldots+\left(b_{88}-b_{87}\right) \cdot 87+\left(20-b_{88}\right) \cdot 88= \\
=20 \cdot 88-\left(b_{1}+b_{2}+\ldots+b_{88}\right), \text { and thus } a_{1}+a_{2}+\ldots+a_{19}+b_{1}+b_{2}+\ldots+b_{88}= \\
=20 \cdot 88=1760 .
\end{gathered}
$$