3. A person adds the page numbers of a book in the order $1,2,3, \cdots$, with one page number being added an extra time, resulting in an incorrect total sum of 2005. Then the page number that was added extra is $\qquad$ .
3.52 pages.

Let the total number of pages in the book be $n$, and the page number that was added extra be $x(1 \leqslant x \leqslant n)$. We have $\frac{n(n+1)}{2}+x=2005$.
And $\frac{n(n+1)}{2}+1 \leqslant 2005 \leqslant \frac{n(n+1)}{2}+n$, which means $n^{2}+n+2 \leqslant 4010 \leqslant n(n+3)$.
Since $\sqrt{4010} \approx 63$, verification shows that the only $n$ satisfying equation (2) is $n=62$.
Substituting into equation (1), we get $x=2005-\frac{62 \times 63}{2}=52$.