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Assume the original book has n pages and the page numbers of the torn page are k and k+1.
Then we have 1 + 2 + 3 + … + n = n(n+1)/2 = 1256 + k + ( k + 1 ).
Since k ≥ 1, we have n(n+1)/2 = 1256 + k + k + 1 ≥ 1256 + 1 + 2 = 1259, n(n+1)/2 ≥ 1259, or n(n+1) ≥ 2518.
Then we have n ≥ 50.
Since n ≥ k, we have n(n+1)/2 ≤ 1256 + k + k + 1 ≤ 1256 + n + n + 1, n(n+1)/2 ≤ 1257 + 2n, n(n+1) ≤ 2514+ 4n, n^2 + n - 4n ≤ 2514, or n^2 – 3n ≤ 2514. Then n ≤ 51.
Case 1) n = 50
We have (50·51)/2 = 1256 + k + (k + 1) or 1275 = 1256 + 2k + 1.
Then 1275 = 1257 + 2k, 2k = 18, and k = 9.
Then, we have k = 9.
Case 2) n = 51
(51·52)/2 = 1256 + k + (k + 1) or 1326 = 1256 + 2k + 1. Then 1326 = 1257 + 2k, 2k = 69, and k = 69/2.
Then, we have k = 69/2, which is not an integer.

Thus, the page numbers of the torn page are 9 and 10.

Therefore, A is the answer.
Answer: A