Find the sum of all the three-digit numbers which are equal to 22 times the sum of their digits.

Let's denote the three-digit number as $\overline{abc}$, where $a, b,$ and $c$ are its digits. This number can be expressed as $100a + 10b + c$.

According to the problem, this number is equal to 22 times the sum of its digits. Therefore, we have the equation: $100a + 10b + c = 22(a + b + c)$

First, let's expand and simplify this equation: $100a + 10b + c = 22a + 22b + 22c$

Next, we move all terms involving $a, b,$ and $c$ to one side of the equation: $100a + 10b + c - 22a - 22b - 22c = 0$

Combine like terms: $78a - 12b - 21c = 0$

To simplify, we can divide the entire equation by 3: $26a - 4b - 7c = 0$

Now, we need to find the values of $a, b,$ and $c$ that satisfy this equation, where $a, b,$ and $c$ are digits (i.e., integers from 0 to 9) and $a$ is a non-zero digit since it's a three-digit number.

Let's solve for $c$ in terms of $a$ and $b$: $26a = 4b + 7c$ $7c = 26a - 4b$ $c = \frac{26a - 4b}{7}$

For $c$ to be a digit, $26a - 4b$ must be divisible by 7. We will test values of $a$ from 1 to 9 and find corresponding $b$ and $c$ that are digits.

1. For $a = 1$: $26(1) - 4b = 7c$ $26 - 4b = 7c$ We need $26 - 4b$ to be divisible by 7. Testing values of $b$:

   • $b = 1$: $26 - 4(1) = 22$ (not divisible by 7)
   • $b = 2$: $26 - 4(2) = 18$ (not divisible by 7)
   • $b = 3$: $26 - 4(3) = 14$ (divisible by 7) $c = \frac{14}{7} = 2$ So, one solution is $a = 1, b = 3, c = 2$. The number is 132.

2. For $a = 2$: $26(2) - 4b = 7c$ $52 - 4b = 7c$ We need $52 - 4b$ to be divisible by 7. Testing values of $b$:

   • $b = 1$: $52 - 4(1) = 48$ (not divisible by 7)
   • $b = 2$: $52 - 4(2) = 44$ (not divisible by 7)
   • $b = 3$: $52 - 4(3) = 40$ (not divisible by 7)
   • $b = 4$: $52 - 4(4) = 36$ (not divisible by 7)
   • $b = 5$: $52 - 4(5) = 32$ (not divisible by 7)
   • $b = 6$: $52 - 4(6) = 28$ (divisible by 7) $c = \frac{28}{7} = 4$ So, another solution is $a = 2, b = 6, c = 4$. The number is 264.

3. For $a = 3$: $26(3) - 4b = 7c$ $78 - 4b = 7c$ We need $78 - 4b$ to be divisible by 7. Testing values of $b$:

   • $b = 1$: $78 - 4(1) = 74$ (not divisible by 7)
   • $b = 2$: $78 - 4(2) = 70$ (divisible by 7) $c = \frac{70}{7} = 10$ (not a valid digit)

Continuing this process for $a = 4$ to $a = 9$, we find that no other values of $a, b,$ and $c$ satisfy the equation with $c$ being a valid digit.

Thus, the only three-digit numbers that satisfy the given condition are 132 and 264. The sum of these numbers is: $132 + 264 = 396$

Therefore, the sum of all the three-digit numbers which are equal to 22 times the sum of their digits is $\boxed{396}$.