Counting ten-digit numbers whose digits are all different and that are divisible by $11111$

The comments indicate that since $9\nmid11111$ yet the sum of digits of an interesting number is divisible by $45$, all interesting numbers are divisible by $99999$; it is then not hard to show that in such numbers the last five digits are complements ($x\to9-x$) of the first five.

The complementary pairs are $09,18,27,36,45$, so there are $5!\cdot2^5$ ways of assigning pairs to the first five positions of an interesting number and then choosing explicitly the digits that go there – except that $4!\cdot2^4$ must be subtracted for those choices giving a leading zero. This leaves $3456$ interesting numbers.

Another way to obtain the same is the following. We have to count the number of ways to fill first $5$ digits of the number only, the second half is (digit-wise) complement of it and is uniquely determined by the first half. So in the following $10$-digit number $$ABCDE - A'B'C'D'E'$$ there are $9$ ways to pick $A$ (excluding $0$), $8$ ways to choose $B$ (since $A$ and $A'$ are filled), $6$ ways for $C$, $4$ for $D$ and $2$ for $E$. Required number is $$9 \times 8 \times 6 \times 4 \times 2 = 3456$$