A positive integer is [i]happy[/i] if:

1. All its digits are different and not $0$,
2. One of its digits is equal to the sum of the other digits.

For example, 253 is a [i]happy[/i] number. How many [i]happy[/i] numbers are there?
To determine how many *happy* numbers there are, we need to count the number of positive integers that satisfy the given conditions:

1. All digits are different and not $0$.
2. One of its digits is equal to the sum of the other digits.

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

1. **Identify the possible digits:**
   Since all digits must be different and not $0$, the possible digits are $1, 2, 3, 4, 5, 6, 7, 8, 9$.

2. **Formulate the condition:**
   For a number to be *happy*, one of its digits must be equal to the sum of the other digits. Let's denote the digits of the number as $d_1, d_2, \ldots, d_n$. Without loss of generality, assume $d_1$ is the digit that equals the sum of the other digits:
   \[
   d_1 = d_2 + d_3 + \ldots + d_n
   \]

3. **Count the valid combinations:**
   We need to count the number of valid combinations for each possible value of $d_1$.

   - **For $d_1 = 3$:**
     The possible combinations are $(3, 1, 2)$ and $(3, 2, 1)$. There are $2$ combinations.

   - **For $d_1 = 4$:**
     The possible combinations are $(4, 1, 3)$, $(4, 3, 1)$, $(4, 2, 2)$. However, $(4, 2, 2)$ is not valid since the digits must be different. Thus, there are $2$ combinations.

   - **For $d_1 = 5$:**
     The possible combinations are $(5, 1, 4)$, $(5, 4, 1)$, $(5, 2, 3)$, $(5, 3, 2)$. There are $4$ combinations.

   - **For $d_1 = 6$:**
     The possible combinations are $(6, 1, 5)$, $(6, 5, 1)$, $(6, 2, 4)$, $(6, 4, 2)$, $(6, 3, 3)$. However, $(6, 3, 3)$ is not valid since the digits must be different. Thus, there are $4$ combinations.

   - **For $d_1 = 7$:**
     The possible combinations are $(7, 1, 6)$, $(7, 6, 1)$, $(7, 2, 5)$, $(7, 5, 2)$, $(7, 3, 4)$, $(7, 4, 3)$. There are $6$ combinations.

   - **For $d_1 = 8$:**
     The possible combinations are $(8, 1, 7)$, $(8, 7, 1)$, $(8, 2, 6)$, $(8, 6, 2)$, $(8, 3, 5)$, $(8, 5, 3)$, $(8, 4, 4)$. However, $(8, 4, 4)$ is not valid since the digits must be different. Thus, there are $6$ combinations.

   - **For $d_1 = 9$:**
     The possible combinations are $(9, 1, 8)$, $(9, 8, 1)$, $(9, 2, 7)$, $(9, 7, 2)$, $(9, 3, 6)$, $(9, 6, 3)$, $(9, 4, 5)$, $(9, 5, 4)$. There are $8$ combinations.

4. **Sum the valid combinations:**
   \[
   2 + 2 + 4 + 4 + 6 + 6 + 8 = 32
   \]

The final answer is $\boxed{32}$.