Problem 1919

There is exactly one solution. The notebook had $64$ pages and the sheet with the page numbers $29$ and $30$ is ripped out.

Let $b > 0$ be the number of sheets. The number of pages will be $2b$. We are looking for a number $2b$ such that
$$1 + 2 + \cdots + (2b - 1) + 2b = \frac{(2b) \cdot (2b + 1)}{2} > 2021.$$
Since $\frac{60^2}{2} = 1800$ has the right order of magnitude, we check the integers beginning with $b = 30$ and find:
$$\frac{62 \cdot 63}{2} = 1953 < 2021 < \frac{64 \cdot 65}{2} = 2080.$$
Therefore, the smallest possible number of pages is $64$.

The torn out sheet $h$ contains the page numbers $2h - 1$ and $2h$ (first odd, then even).
This gives the equation
$$2h - 1 + 2h = 2080 - 2021 = 59$$
which implies $h = 15$.

So, one solution is that the book originally had $32$ sheets and the $15$th sheet with page numbers $29$ and $30$ has been torn out.

It remains to explain why this is the only solution. If the book has $64$ sheets, this is the only possibility because $h$ could be computed uniquely. Now, assume that the number $b$ of sheets is larger than $32$ and the number of pages at least $66$.

The sheet that has been torn out can have at most page numbers $2b - 1$ and $2b$, so the remaining sum is at least $1 + 2 + \cdots + 63 + 64 = 2080 > 2021$. So there cannot be a solution with more than $32$ sheets.