3. A book has a total of 61 pages, sequentially numbered as 1, 2, ..., 61. Someone, while adding these numbers, mistakenly reversed the digits of two two-digit page numbers (a two-digit number of the form $\overline{a b}$ was treated as $\overline{b a}$), resulting in a total sum of 2008. Therefore, the maximum sum of these two two-digit page numbers on the book is $\qquad$

Translating the problem statement into English while preserving the original formatting and line breaks, the result is as follows:

3. A book has a total of 61 pages, sequentially numbered as 1, 2, ..., 61. Someone, while adding these numbers, mistakenly reversed the digits of two two-digit page numbers (a two-digit number of the form $\overline{a b}$ was treated as $\overline{b a}$), resulting in a total sum of 2008. Therefore, the maximum sum of these two two-digit page numbers on the book is $\qquad$
3.68.

Notice that $1+2+\cdots+61=1891$,
$$
2008-1891=117 \text {. }
$$

Since the page numbers in the form of $\overline{a b}$ are read as $\overline{b a}$, the difference in the sum will be $9|a-b|$, because $a$ and $b$ can only take values from 1,2, $\cdots, 9$, $|a-b| \leqslant 8$, so,
$9|a-b| \leqslant 72$.
Since $117=72+45=63+54$, let the two numbers that were misread be $\overline{a b}$ and $\overline{c d}$.

If $9|a-b|=72,9|c-d|=45$, then only $\overline{a b}=19$, and $\overline{c d}$ can be $16,27,38,49$, in this case, the maximum value of $\overline{a b}+$ $\overline{c d}$ is 68;

If $9|a-b|=63,9|c-d|=54$, then $\overline{a b}$ can be 18,29, and $\overline{c d}$ can be $17,28,39$, in this case, the maximum value of $\overline{a b}+\overline{c d}$ is also 68.