Let $N$ be the greatest integer multiple of $8,$ no two of whose digits are the same. What is the remainder when $N$ is divided by $1000?$
1. **Understanding the Problem:**
   We need to find the greatest integer multiple of 8, no two of whose digits are the same, and determine the remainder when this number is divided by 1000.

2. **Finding the Greatest Integer Multiple of 8:**
   To maximize the integer, we should use the largest possible digits without repetition. The largest digits are 9, 8, 7, 6, 5, 4, 3, 2, 1, and 0.

3. **Constructing the Largest Number:**
   The largest number using these digits is 9876543210. We need to check if this number is a multiple of 8.

4. **Divisibility Rule for 8:**
   A number is divisible by 8 if its last three digits form a number that is divisible by 8. The last three digits of 9876543210 are 210.

5. **Checking Divisibility of 210 by 8:**
   \[
   210 \div 8 = 26.25
   \]
   Since 210 is not divisible by 8, we need to adjust the last three digits to form a number that is divisible by 8.

6. **Finding the Largest Permutation of Last Three Digits Divisible by 8:**
   We need to find the largest permutation of the digits 0, 1, 2 that is divisible by 8. The permutations are 012, 021, 102, 120, 201, and 210. Among these, 120 is divisible by 8:
   \[
   120 \div 8 = 15
   \]
   Therefore, 120 is the largest permutation of 0, 1, 2 that is divisible by 8.

7. **Forming the Largest Number:**
   Replace the last three digits of 9876543210 with 120 to get 9876543120.

8. **Finding the Remainder When Divided by 1000:**
   The last three digits of 9876543120 are 120. Therefore, the remainder when 9876543120 is divided by 1000 is 120.

The final answer is $\boxed{120}$