A leaf is torn from a paperback novel. The sum of the numbers on the remaining pages is $15000$. What are the page numbers on the torn leaf?
1. Let \( n \) be the total number of pages in the book. The 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. Suppose pages \( k-1 \) and \( k \) are torn out. The sum of the remaining pages is:
   \[
   S' = S - (k-1) - k = \frac{n(n+1)}{2} - (2k-1)
   \]
   Given that the sum of the remaining pages is 15000, we have:
   \[
   \frac{n(n+1)}{2} - (2k-1) = 15000
   \]
3. Rearrange the equation to solve for \( n \):
   \[
   \frac{n(n+1)}{2} - 2k + 1 = 15000
   \]
   \[
   \frac{n(n+1)}{2} = 15000 + 2k - 1
   \]
   \[
   \frac{n(n+1)}{2} = 14999 + 2k
   \]
   \[
   n(n+1) = 2(14999 + 2k)
   \]
   \[
   n(n+1) = 29998 + 4k
   \]
4. To find the range of \( n \), we solve the quadratic inequality:
   \[
   n(n+1) \approx 30000
   \]
   Solving \( n^2 + n - 30000 = 0 \) using the quadratic formula:
   \[
   n = \frac{-1 \pm \sqrt{1 + 4 \cdot 30000}}{2} = \frac{-1 \pm \sqrt{120001}}{2}
   \]
   Approximating the square root:
   \[
   \sqrt{120001} \approx 346.41
   \]
   \[
   n \approx \frac{-1 + 346.41}{2} \approx 172.70
   \]
   Since \( n \) must be an integer, we check \( n = 173 \) and \( n = 174 \).

5. For \( n = 173 \):
   \[
   173 \cdot 174 = 30002
   \]
   \[
   30002 = 29998 + 4k
   \]
   \[
   4k = 4
   \]
   \[
   k = 1
   \]
   This is not possible since \( k \) must be greater than 1.

6. For \( n = 174 \):
   \[
   174 \cdot 175 = 30450
   \]
   \[
   30450 = 29998 + 4k
   \]
   \[
   4k = 452
   \]
   \[
   k = 113
   \]
   Therefore, the torn pages are \( 112 \) and \( 113 \).

The final answer is \(\boxed{(112, 113)}\).