9-digit number

Let $x$ be the sum of digits in the odd places and $y$ be the sum of digits in the even places. Note that

(i) $\;\;\;\;x \ge y$ as $N$ is the largest such number.

(i) $\;\;\;\;x-y$ is divisible by 11 as $N$ is divisible by 11.

(ii) $\;\;\;\ 10=1+2+3+4 \le y \le x \le 9+8+7+6+5 =35$ . This means that $x-y \le 25$

(iii) $\;\;\;\ x + y$ is the sum of all integers from 1 to 9, which is 45, an odd number. This means that $x - y = (x + y ) - 2y$ is an odd number too.

From (i) to (iv), we have $x-y =11$ . This means that $x= 28,\;\; y =17$ .

As $N$ is the largest, consider the first few digits of $N= 98765****$ . Since $9+7+5=21$ , we need to choose 2 digits from $\{1, 2, 3, 4\}$ , whose sum is 7, so that $x=28$ . Clearly, we need 3 and 4. (As 3 and 4 had been chosen, we left with 1 and 2 for y. Now $y=8+6+1+2=17$ , yeah!!!)

Then $N=987652413$ and the desired last 4 digit is $\textcolor{#D61F06} {2413}$ .

Let the number be $\overline {abcdefghi}$ with $a\neq b\neq c\neq d\neq e\neq f\neq g\neq h\neq i$ .

Then $(a+c+e+g+i)-(b+d+f+h)$ must be divisible by $11$ .

The two sums can't be equal, since the sum of the digits of the number is $45$ which is odd.

For similar reason, they can't differ by $22$ also.

Since all the digits of the number are distinct, therefore the difference can't be divisible by higher multiples of $11$ than $2$ .

So the only possibility is that the difference is $11$ .

There are different groupings of the numbers $1$ through $9$ satisfying this condition, viz.

$(a, c, e, g, i) =(3,4,6,7,8), (b, d, f, h) =(1,2,5,9); (a, c, e, g, i) =(2,5,6,7,8), (b, d, f, h) =(1,3,4,9)$ and so on.

For the first group, there are several numbers again :

$897561423, 897562413, 798562413, 798561423, 817265493$ and so on.

The largest of them is $\boxed {987652413}$ , whose last four digits are $\boxed {2413}$ .