A [i]trifecta[/i] is an ordered triple of positive integers $(a, b, c)$ with $a < b < c$ such that $a$ divides $b$, $b$ divides $c$, and $c$ divides $ab$. What is the largest possible sum $a + b + c$ over all trifectas of three-digit integers?
To solve this problem, we need to find the largest possible sum \(a + b + c\) for a trifecta \((a, b, c)\) of three-digit integers such that:
1. \(a < b < c\)
2. \(a\) divides \(b\)
3. \(b\) divides \(c\)
4. \(c\) divides \(ab\)

Let's break down the problem step-by-step:

1. **Identify the constraints:**
   - \(a, b, c\) are three-digit integers.
   - \(a < b < c\)
   - \(a \mid b\) (i.e., \(b = ka\) for some integer \(k\))
   - \(b \mid c\) (i.e., \(c = lb\) for some integer \(l\))
   - \(c \mid ab\) (i.e., \(ab = mc\) for some integer \(m\))

2. **Express \(b\) and \(c\) in terms of \(a\):**
   - Since \(a \mid b\), let \(b = ka\) where \(k\) is an integer greater than 1.
   - Since \(b \mid c\), let \(c = lb = lka\) where \(l\) is an integer greater than 1.

3. **Substitute \(b\) and \(c\) into the condition \(c \mid ab\):**
   - We have \(c = lka\).
   - The condition \(c \mid ab\) becomes \(lka \mid a(ka)\).
   - Simplifying, we get \(lka \mid k a^2\).
   - Since \(a\) is a common factor, we can cancel \(a\) from both sides: \(lk \mid ka\).
   - This simplifies to \(l \mid a\).

4. **Find suitable values for \(a, b, c\):**
   - Since \(a, b, c\) are three-digit integers, \(100 \leq a < b < c \leq 999\).
   - We need to find the largest possible sum \(a + b + c\).

5. **Test possible values:**
   - Let's start with the smallest three-digit integer \(a = 100\).
   - For \(b\) to be a multiple of \(a\), \(b = 2a = 200\).
   - For \(c\) to be a multiple of \(b\), \(c = 2b = 400\).
   - Check if \(c \mid ab\): \(400 \mid 100 \cdot 200 = 20000\), which is true.

   Therefore, \((a, b, c) = (100, 200, 400)\) is a valid trifecta.

6. **Calculate the sum:**
   - The sum \(a + b + c = 100 + 200 + 400 = 700\).

Thus, the largest possible sum \(a + b + c\) over all trifectas of three-digit integers is \(700\).

The final answer is \(\boxed{700}\).