If $m, n$, and $p$ are three different natural numbers, each between $2$ and $9$, what then are all the possible integer value(s) of the expression $\frac{m+n+p}{m+n}$?
1. We start with the given expression $\frac{m+n+p}{m+n}$, where $m, n$, and $p$ are distinct natural numbers between $2$ and $9$.

2. First, we note that $m$ and $n$ are distinct natural numbers between $2$ and $9$. Therefore, the minimum value of $m+n$ is $2+3=5$ (since $m$ and $n$ are distinct), and the maximum value of $m+n$ is $8+9=17$.

3. Next, we consider the range of $p$. Since $p$ is also a natural number between $2$ and $9$, we have $2 \leq p \leq 9$.

4. We now analyze the expression $\frac{m+n+p}{m+n}$:
   \[
   \frac{m+n+p}{m+n} = 1 + \frac{p}{m+n}
   \]

5. To find the possible integer values of $\frac{m+n+p}{m+n}$, we need to consider the range of $\frac{p}{m+n}$:
   \[
   0 < \frac{p}{m+n} < 2
   \]
   This is because $p$ is a positive number less than $m+n$ (since $m, n, p$ are distinct and $m+n \geq 5$).

6. We need to check if $\frac{p}{m+n}$ can be an integer. Since $p$ and $m+n$ are distinct natural numbers, $\frac{p}{m+n}$ cannot be an integer. Therefore, $\frac{p}{m+n}$ is always a proper fraction.

7. Given that $\frac{p}{m+n}$ is a proper fraction, the expression $1 + \frac{p}{m+n}$ will always be a number between $1$ and $2$ but not equal to $2$.

8. The only integer value within this range is $1$. Therefore, the only possible integer value of $\frac{m+n+p}{m+n}$ is $1$.

The final answer is $\boxed{1}$.