4321
4132
3521
4123
2413
1243

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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 $\textcolor{#D61F06} {2413}$ .