3. Among all the eight-digit numbers formed by the digits $1,2, \cdots, 8$ without repetition, the number of those divisible by 11 is. $\qquad$
3.4608 .

Since there are 4 odd numbers in $1,2, \cdots, 8$, the algebraic sum after arbitrarily adding “+” and “-” signs is always even. Because the difference between the sum of the largest four numbers and the sum of the smallest four numbers in $1,2, \cdots, 8$ is no more than 16, for each eight-digit number that meets the conditions, the sum of the four digits in the odd positions must equal the sum of the four digits in the even positions. Since $1+2+\cdots+8=36$, divide $1,2, \cdots, 8$ into two groups of four numbers each, with the sum of each group being 18, and consider the group containing 8, the sum of the other three numbers in the group is 10, there are only four cases:
$$
(1,2,7,8),(1,3,6,8),(1,4,5,8),(2,3,5,8) \text {. }
$$

For each case, the group containing 8 can be placed in the odd positions or the even positions, resulting in $2 \times 4!\times 4!$ numbers.

In the four cases, a total of $8 \times 4!\times 4!=4608$ eight-digit numbers that meet the conditions are obtained.