For positive integer $n$ let $a_n$ be the integer consisting of $n$ digits of $9$ followed by the digits $488$. For example, $a_3 = 999,488$ and $a_7 = 9,999,999,488$. Find the value of $n$ so that an is divisible by the highest power of $2$.
1. Let's start by expressing \( a_n \) in a more manageable form. Given \( a_n \) is the integer consisting of \( n \) digits of 9 followed by the digits 488, we can write:
   \[
   a_n = 10^n \cdot 999 + 488
   \]
   This is because \( 999 \ldots 999 \) (with \( n \) digits of 9) can be written as \( 10^n - 1 \).

2. Therefore, we have:
   \[
   a_n = 10^n \cdot 999 + 488 = 10^n \cdot (10^3 - 1) + 488 = 10^n \cdot 999 + 488
   \]
   Simplifying further:
   \[
   a_n = 10^n \cdot 1000 - 10^n + 488 = 10^{n+3} - 10^n + 488
   \]

3. To find the highest power of 2 that divides \( a_n \), we need to simplify the expression:
   \[
   a_n = 10^{n+3} - 512
   \]
   Here, \( 512 = 2^9 \).

4. We need to determine the highest power of 2 that divides \( 10^{n+3} - 512 \). We use the fact that \( 10 \equiv 2 \pmod{8} \), so:
   \[
   10^{n+3} \equiv 2^{n+3} \pmod{8}
   \]
   Since \( 2^{n+3} \) is always divisible by 8 for \( n \geq 3 \), we need to consider the power of 2 in \( 10^{n+3} - 512 \).

5. For \( n = 6 \):
   \[
   a_6 = 10^9 - 512
   \]
   We need to check the divisibility of \( 10^9 - 512 \) by higher powers of 2. We know:
   \[
   10^9 = (2 \cdot 5)^9 = 2^9 \cdot 5^9
   \]
   Therefore:
   \[
   10^9 - 512 = 2^9 \cdot 5^9 - 2^9 = 2^9 (5^9 - 1)
   \]

6. Now, we need to check the power of 2 in \( 5^9 - 1 \):
   \[
   5^9 - 1 = (5-1)(5^8 + 5^7 + 5^6 + 5^5 + 5^4 + 5^3 + 5^2 + 5 + 1) = 4 \cdot (5^8 + 5^7 + 5^6 + 5^5 + 5^4 + 5^3 + 5^2 + 5 + 1)
   \]
   Since \( 4 = 2^2 \), we have:
   \[
   5^9 - 1 = 2^2 \cdot k \quad \text{(where \( k \) is an odd number)}
   \]

7. Therefore:
   \[
   10^9 - 512 = 2^9 \cdot 2^2 \cdot k = 2^{11} \cdot k
   \]
   where \( k \) is odd.

8. Hence, the highest power of 2 that divides \( a_6 \) is \( 2^{11} \).

The final answer is \( \boxed{6} \).