Austrian Mathematical Olympiad

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.