Find the number of positive integers $n$ such that the highest power of $7$ dividing $n!$ is $8$.
To find the number of positive integers \( n \) such that the highest power of \( 7 \) dividing \( n! \) is \( 8 \), we need to use the formula for the highest power of a prime \( p \) dividing \( n! \):

\[
\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots
\]

In this case, \( p = 7 \) and we want the sum to be \( 8 \). Therefore, we need:

\[
\left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor = 8
\]

1. **Finding the smallest \( n \):**

   We start by finding the smallest \( n \) such that the sum of the floor functions equals \( 8 \).

   \[
   \left\lfloor \frac{n}{7} \right\rfloor + \left\lfloor \frac{n}{49} \right\rfloor = 8
   \]

   Let \( n = 49k + r \) where \( 0 \leq r < 49 \). Then,

   \[
   \left\lfloor \frac{49k + r}{7} \right\rfloor = 7k + \left\lfloor \frac{r}{7} \right\rfloor
   \]

   and

   \[
   \left\lfloor \frac{49k + r}{49} \right\rfloor = k
   \]

   Therefore,

   \[
   7k + \left\lfloor \frac{r}{7} \right\rfloor + k = 8 \implies 8k + \left\lfloor \frac{r}{7} \right\rfloor = 8
   \]

   This implies \( k = 1 \) and \( \left\lfloor \frac{r}{7} \right\rfloor = 0 \), so \( 0 \leq r < 7 \). The smallest \( n \) is:

   \[
   n = 49 \cdot 1 + 0 = 49
   \]

2. **Finding the largest \( n \):**

   Next, we find the largest \( n \) such that the sum of the floor functions equals \( 8 \).

   \[
   8k + \left\lfloor \frac{r}{7} \right\rfloor = 8
   \]

   This implies \( k = 1 \) and \( \left\lfloor \frac{r}{7} \right\rfloor = 0 \), so \( 0 \leq r < 7 \). The largest \( n \) is:

   \[
   n = 49 \cdot 1 + 6 = 55
   \]

3. **Counting the integers:**

   The integers \( n \) that satisfy the condition are from \( 49 \) to \( 55 \), inclusive. Therefore, the number of such integers is:

   \[
   55 - 49 + 1 = 7
   \]

The final answer is \(\boxed{7}\).