An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of original number, the resulting number is divisible by 5. After deleting three digits from the right of original number, the resulting number is divisible by 3. After deleting five digits from the right of original number, the resulting number is divisible by 2. Which of the following is a possible value for the sum of the middle three digits of the number?

The correct option is 8.

Let the 7-digit Identity Card number be represented as ABCDEFG, where each letter represents a distinct digit from the set:
{1, 2, 4, 5, 7, 8, 9}
We are given various divisibility conditions after successively removing digits from the right. Let's analyze these conditions step-by-step:

1. Five digits deleted from the right:
After deleting five digits from the right, we are left with the 2-digit number AB. We are told AB is divisible by 2.
For a number to be divisible by 2, its last digit must be even. In our set of digits {1, 2, 4, 5, 7, 8, 9}, the even digits are 2, 4, and 8.
Therefore, B must be one of {2, 4, 8}.

2. Three digits deleted from the right:
After deleting three digits from the right, we are left with the 4-digit number ABCD. We are told ABCD is divisible by 3.
For a number to be divisible by 3, the sum of its digits must be a multiple of 3.
Therefore:
$A+B+C+D$ is a multiple of 3.

3. Two digits deleted from the right:
After deleting two digits from the right, we are left with the 5-digit number ABCDE. We are told ABCDE is divisible by 5.
For a number to be divisible by 5, its last digit must be 0 or 5. Since our set of digits only contains {1, 2, 4, 5, 7, 8, 9}, E must be 5.
Thus, E = 5.

4. One digit deleted from the right:
After deleting one digit from the right, we are left with the 6-digit number ABCDEF. We are told ABCDEF is divisible by 6.
For a number to be divisible by 6, it must be divisible by both 2 and 3.
Since ABCDEF is divisible by 2, its last digit F must be even.
Thus, F must be one of {2, 4, 8}.
Since ABCDEF is divisible by 3, the sum of its digits must be a multiple of 3:
$A+B+C+D+E+F$ is a multiple of 3.

5. The original 7-digit number ABCDEFG:
We are told the original number is divisible by 9.
The sum of all seven digits (A + B + C + D + E + F + G) must be a multiple of 9.
Since the digits are a permutation of {1, 2, 4, 5, 7, 8, 9}, the sum of the digits is:
$1+2+4+5+7+8+9=36$
Since 36 is divisible by 9, this condition is automatically satisfied for any arrangement of these seven digits.

Determining the digits:
We know that B and F must be even, chosen from {2, 4, 8}.
Let's look at the divisibility of ABCD and ABCDEF by 3:
Let $S_4=A+B+C+D$ (must be a multiple of 3).
Let $S_6=A+B+C+D+E+F=S_4+E+F$ (must be a multiple of 3).
Since $S_4$ is a multiple of 3, $E+F$ must also be a multiple of 3.
We know E = 5. Thus, $5+F$ must be a multiple of 3.
Since F can only be 2, 4, or 8:
- If F = 2, $5+2=7$ (not a multiple of 3)
- If F = 4, $5+4=9$ (multiple of 3) ⇒ F = 4
- If F = 8, $5+8=13$ (not a multiple of 3)
Thus, we must have F = 4.

Since F = 4 and B is also even, B must be one of {2, 8}.

The middle three digits of the 7-digit number ABCDEFG are C, D, and E.
We need to find the sum of these middle three digits: $C+D+E$.
Since E = 5, the sum is $C+D+5$.
Let's check the divisibility of ABCD by 3:
$A+B+C+D=3k$ for some integer k.
The sum of all digits is:
$(A+B+C+D)+E+F+G=36$
Substitute the known values and relations:
$3k+5+4+G=36$
$3k+9+G=36$
$3k+G=27$
Since 27 and 3k are multiples of 3, G must be a multiple of 3.
The only digit in our set {1, 2, 4, 5, 7, 8, 9} that is a multiple of 3 is 9.
Thus, G = 9.

Now we have:
- E = 5
- F = 4
- G = 9
The remaining digits for {A, B, C, D} are {1, 2, 7, 8}.
Since B must be even, B must be either 2 or 8.
Let's test both possibilities to find a valid combination where $A+B+C+D$ is a multiple of 3 (which is guaranteed since $A+B+C+D=36-5-4-9=18$, a multiple of 3).
The sum of the middle three digits is:
$\text{Sum}=C+D+E=C+D+5$
Since {A, B, C, D} = {1, 2, 7, 8}:
- If B = 8, then {A, C, D} must be a permutation of {1, 2, 7}.
Then the two digits C and D are chosen from {1, 2, 7}.
Possible values for C + D from {1, 2, 7}:
- $1+2=3\Rightarrow \text{Sum}=3+5=8$
- $1+7=8\Rightarrow \text{Sum}=8+5=13$
- $2+7=9\Rightarrow \text{Sum}=9+5=14$

Since 8 is one of the choices listed in the options, a sum of 8 is indeed a possible value.