The pages of a notebook are numbered consecutively so that the numbers $1$ and $2$ are on the second sheet, numbers $3$ and $4$, and so on. A sheet is torn out of this notebook. All of the remaining page numbers are addedand have sum $2021$.
(a) How many pages could the notebook originally have been?
(b) What page numbers can be on the torn sheet?

(Walther Janous)
1. Let the number of pages in the notebook be \( n \). The total sum of the page numbers from 1 to \( n \) is given by the formula for the sum of the first \( n \) natural numbers:
   \[
   S = \frac{n(n+1)}{2}
   \]
2. Let the smaller number on the torn sheet be \( k \). Since each sheet contains two consecutive numbers, the numbers on the torn sheet are \( k \) and \( k+1 \). The sum of these two numbers is:
   \[
   k + (k+1) = 2k + 1
   \]
3. The sum of the remaining page numbers after tearing out the sheet is given as 2021. Therefore, we have:
   \[
   \frac{n(n+1)}{2} - (2k + 1) = 2021
   \]
4. Rearrange the equation to solve for \( n \):
   \[
   \frac{n(n+1)}{2} = 2021 + 2k + 1
   \]
   \[
   \frac{n(n+1)}{2} = 2022 + 2k
   \]
   \[
   n(n+1) = 2(2022 + 2k)
   \]
   \[
   n(n+1) = 4044 + 4k
   \]
5. For \( n(n+1) = 4044 + 4k \) to hold, \( n \) or \( n+1 \) must be divisible by 4. This implies that \( n \) must be congruent to 0 or 3 modulo 4. Therefore, the possible values of \( n \) are any integers \( \geq 64 \) that are congruent to 0 or 3 modulo 4.

6. To find the possible values of \( k \), we need to ensure that \( n^2 + n - (4044 + 4k) = 0 \) has positive integer roots. Using the quadratic formula:
   \[
   n = \frac{-1 \pm \sqrt{1 + 4(4044 + 4k)}}{2}
   \]
   \[
   n = \frac{-1 \pm \sqrt{1 + 16176 + 16k}}{2}
   \]
   \[
   n = \frac{-1 \pm \sqrt{16177 + 16k}}{2}
   \]
7. For \( n \) to be an integer, \( 16177 + 16k \) must be a perfect square. Let \( 16177 + 16k = m^2 \) for some integer \( m \):
   \[
   16k = m^2 - 16177
   \]
   \[
   k = \frac{m^2 - 16177}{16}
   \]
8. \( m^2 - 16177 \) must be divisible by 16. Since \( 16177 \equiv 1 \pmod{16} \), we have:
   \[
   m^2 \equiv 1 \pmod{16}
   \]
   This implies \( m \equiv 1 \) or \( m \equiv 15 \pmod{16} \).

9. Therefore, \( m \) must be an odd number greater than \( \sqrt{16177} \approx 127.2 \). The smallest odd number greater than 127.2 is 129. So, \( m \) can be 129, 145, 161, etc.

10. For each \( m \), we can calculate \( k \):
    \[
    k = \frac{m^2 - 16177}{16}
    \]
    For example, if \( m = 129 \):
    \[
    k = \frac{129^2 - 16177}{16} = \frac{16641 - 16177}{16} = \frac{464}{16} = 29
    \]

Conclusion:
The possible values of \( n \) are integers \( \geq 64 \) that are congruent to 0 or 3 modulo 4. The possible values of \( k \) are values that give positive integer roots to the equation \( n^2 + n - (4044 + 4k) = 0 \), where \( k = \frac{m^2 - 16177}{16} \) and \( m \) is an odd number greater than 127.2.

The final answer is \( \boxed{ n } \) must be congruent to 0 or 3 modulo 4 and \( k = \frac{m^2 - 16177}{16} \) where \( m \) is an odd number greater than 127.2.