Solution:

The digits sum to $36.$ Divisibility by $11$ requires the alternating sum of digits to be a multiple of $11,$ so if the four digits in odd positions sum to $a,$ then $a - (36 - a) = 2a - 36$ must be a multiple of $11.$ Since $10 \le a \le 26,$ the only possibility is $a = 18:$ each block of four positions carries digit sum $18.$ The four-element subsets of $\{1, \ldots, 8\}$ with sum $18$ are $$\{1,2,7,8\},\ \{1,3,6,8\},\ \{1,4,5,8\},\ \{1,4,6,7\},\ \{2,3,5,8\},\ \{2,3,6,7\},\ \{2,4,5,7\},\ \{3,4,5,6\},$$ eight in all, and they come in complementary pairs.

Choose which of the $8$ subsets occupies the even positions (which include the units place); the complement fills the odd positions. If that subset contains $k$ of the even digits, then the units digit can be chosen in $k$ ways, the rest of the even positions in $3!$ ways, and the odd positions in $4!$ ways, for $144k$ numbers. Complementary subsets have $k$-values summing to $4,$ so over all $8$ choices $\sum k = 16.$ Hence $N = 144 \cdot 16 = 2304,$ and $N - 2025 = 279.$