Give it your best shot

The divisibility test for $11$ is that the difference of the sum of digits at odd places $(1,3,5,...)$ and even places $(2,4,6,...)$ must be zero or a multiple of $11$ .

$O_{1}-E_{1}-O_{2}-E_{2}-O_{3}-E_{3}-O_{4}-E_{4}-O_{5}$

Let the $O$ represent odd place numbers and $E$ represent even place numbers. We can use the numbers $0,1,2,3,.....8,9$ , so let us first suppose the maximmum possible number we can make with these digits each being used only once - $987654321$ .

As the question requires that we need to make the maximum possible number divisible by $11$ in that way, so we will try replacing out numbers from the right with other digits so that the number on a whole becomes divisible by $11$ .

As in $987654321$ difference between odd and even places is $5$ , so we can make it $11$ or $0$ .

Replacing ones digit (first digit);won't work because we can replace it only with $0$ which doesn't make the situation better.

Replacing first two digits (ones and tens) would again clearly not work as we have only two digits to with here ( $0$ and $1$ ) which don't make any difference.

Replacing first three digits : Only integers we can use - $0,1,2$ . Try it out not going to make any difference and make the number divisible by $11$ .

Now, comes the turn to replace first four digits, with a bit of logic and trial we can actually make the difference of the odd places numbers and the even places number $11$ and thus making the number divisible by $11$ , by replacing the last four digits as $2 - 4 - 1 - 3$ .

And we got the answer - $987652413$