Let the number of pages on the notebook be $n$. Let the smaller number on the torn sheet be $k$. So $\frac{n(n+1)}{2}-(2k+1)=2021$, or $n(n+1)=4044+4k$, so $n$ or $n+1$ must be divisible by $4$. Thus the possible values of $n$ are any integers $64$ or greater that are congruent to $3$ or $0\mod 4$. The possible values of $k$ are values that give positive integer roots to the equation $n^2+n-(4044+4k)$. By quadratic formula $\sqrt{4045+4k}$ must be an odd square, so the possible values of k are $\frac{s-4045}{4}$, where $s$ is any odd perfect square greater than $4045$.