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because of the word "ANY" , we come to know that if the number is divisible by 11, and because ANY permutation is divisible by 11 , ALL the digits MUST be the same , and this thing happens only for a number with EVEN number of digits. As the number is divisible by 5 , it's unit digit can be 0 or 5 , but as all digits are same and number is 8-digit , we conclude that the digit is not 0 , it's 5..... Thus the number given to us indirectly is $55555555$ and sum of the digits of ALL possible numbers here means only the number itself . , hence answer is $\boxed{40}$

The proof of the above used fact can be done by

Let the number be $\overline{abcdefgh}$ then we have this divisible by 11 , and also $\overline{abcdefhg}$ divisible by 11. Thus remainder of $gh$ is same as $hg$ mod 11. And h=0 or 5. Thus we can surely say that gh=00 or 55 . And doing this for all possible pairs of numbers, we get that all the digits are 0 or 5. But if 0 then number is not 8 digit hence the number is $\Huge{55555555}$ Thus it gives digit sum of $\Huge{\color{limegreen}{40}}$

We claim that the only number satisfying it is $55555555$ .

Suppose (for the sake of contradiction) that the number has two different digits. Then the number also has two adjacent digits that are different, since otherwise any pair of adjacent digits are equal and hence all digits are equal, contradiction.

Suppose our number is $N = \overline{\ldots ab \ldots}$ . Denote the other digits of the number as a single number $n$ , so that our number is $N = n + 10^{k+1}a + 10^kb$ for some $k$ . Observe that $10^{k+1} = 10 \cdot 10^k \equiv -10^k \pmod{11}$ , so $N \equiv n + 10^k(b-a) \pmod{11}$ . And since this is divisible by $11$ , we need $n + 10^k(b-a) \equiv 0 \pmod{11}$ .

Now consider a permutation of the number, where we swap $a$ and $b$ and leaves everything else intact. Then our new number is $N' = \overline{\ldots ba \ldots} = n + 10^{k+1}b + 10^ka \equiv n + 10^k(a-b) \pmod{11}$ . And since this is also divisible by $11$ , we need $n + 10^k(a-b) \equiv 0 \pmod{11}$ .

Thus $n + 10^k(b-a) \equiv n + 10^k(a-b) \pmod{11}$ , or $2 \cdot 10^k(b-a) \equiv 0 \pmod{11}$ . Since $2 \cdot 10^k$ is always relatively prime to $11$ , it has a multiplicative inverse $(2 \cdot 10^k)^{-1}$ modulo $11$ . Multiplying both sides with it gives $(2 \cdot 10^k)^{-1}(2 \cdot 10^k) \cdot (b-a) \equiv (2 \cdot 10^k)^{-1} \cdot 0 \pmod{11}$ , or $b-a \equiv 0 \pmod{11}$ .

But $a,b$ are digits; in particular, $0 \le a,b \le 9$ . If $a \neq b$ , due to $b-a \equiv 0 \pmod{11}$ we need $|b-a| \ge 11$ (since the quantity $b-a$ is divisible by $11$ but is not equal to $0$ ). But $0-9 \le b-a \le 9-0$ , or $|b-a| \le 9$ , contradiction. Thus our assumption that there are two different digits in the number can't hold, and thus all digits in the number are equal.

Since the number is divisible by $5$ , the last digit must be $0$ or $5$ . If the last digit is $0$ , this forces the number to be $00000000$ which is not a 8-digit number. So the number is $55555555$ . The only permutation of this number is itself, and thus the sum of digit sums is simply the digit sum of $55555555$ , which is $5+5+5+5+5+5+5+5 = \boxed{40}$ .

It is a property of multiples of 11 that the alternating sum of the digits equals zero. By this I mean, if the number $\overline{abcdefgh}$ is a multiple of 11, then $a - b + c - d + e - f + g - h = 0$ . Alternatively $a + c + e + g = b + d + f + h.$ We are given that any permutation of the digits is also a multiple of 11. Thus $\overline{abcdefhg}$ is also a multiple of 11, so $a + c + e + h = b + d + f + g.$ These two equations imply that $g = h$ . This equality holds for any pair of digits, thus the eight-digit number has eight identical digits. We are told that it is a multiple of 5 (and that 00000000 is not considered an eight-digit number), thus the only number satisfying the criteria is 55555555. Its sum of digits is 40.