We have a list of N consecutive 3-digit numbers, each of which is not divisible by its digit sum. What is the largest possible value of N?

To find the largest possible value of N, we need to look for consecutive 3-digit numbers that are not divisible by their digit sum. Let's break down the problem into steps:

Step 1: Identify the consecutive 3-digit numbers.
We know that a 3-digit number ranges from 100 to 999. So, we have a total of 999 - 100 + 1 = 900 consecutive 3-digit numbers.

Step 2: Determine the digit sum of each number.
For example, let's take the number 123. The digit sum is calculated by adding each digit: 1 + 2 + 3 = 6.

Step 3: Check if each number is divisible by its digit sum.
If a number is divisible by its digit sum, it means the number can be evenly divided by the sum of its digits with no remainder.

Now, let's consider the largest possible value of N:

Starting from 100, we can check each consecutive 3-digit number to see if it is divisible by its digit sum.

100: Digit sum = 1 + 0 + 0 = 1. 100 is not divisible by 1.
101: Digit sum = 1 + 0 + 1 = 2. 101 is not divisible by 2.
102: Digit sum = 1 + 0 + 2 = 3. 102 is not divisible by 3.
... continuing with the sequence ...

999: Digit sum = 9 + 9 + 9 = 27. 999 is not divisible by 27.

We have checked all 900 numbers and found that none of them are divisible by their digit sum.

Therefore, the largest possible value of N is 900 since all N consecutive 3-digit numbers are not divisible by their digit sum.