For which integer $n$ is $N = 20^n + 16^n - 3^n - 1$ divisible by $323$?
To determine for which integer \( n \) the expression \( N = 20^n + 16^n - 3^n - 1 \) is divisible by \( 323 \), we need to check the divisibility by both \( 19 \) and \( 17 \) since \( 323 = 19 \times 17 \).

1. **Divisibility by 19:**
   - We need \( 20^n + 16^n - 3^n - 1 \equiv 0 \pmod{19} \).
   - Note that \( 20 \equiv 1 \pmod{19} \), so \( 20^n \equiv 1^n \equiv 1 \pmod{19} \).
   - Also, \( 16 \equiv -3 \pmod{19} \), so \( 16^n \equiv (-3)^n \pmod{19} \).
   - Therefore, the expression becomes:
     \[
     1 + (-3)^n - 3^n - 1 \equiv 0 \pmod{19}
     \]
   - Simplifying, we get:
     \[
     (-3)^n - 3^n \equiv 0 \pmod{19}
     \]
   - This implies:
     \[
     (-3)^n \equiv 3^n \pmod{19}
     \]
   - For this to hold, \( n \) must be even because \( (-3)^n = 3^n \) if and only if \( n \) is even.

2. **Divisibility by 17:**
   - We need \( 20^n + 16^n - 3^n - 1 \equiv 0 \pmod{17} \).
   - Note that \( 20 \equiv 3 \pmod{17} \), so \( 20^n \equiv 3^n \pmod{17} \).
   - Also, \( 16 \equiv -1 \pmod{17} \), so \( 16^n \equiv (-1)^n \pmod{17} \).
   - Therefore, the expression becomes:
     \[
     3^n + (-1)^n - 3^n - 1 \equiv 0 \pmod{17}
     \]
   - Simplifying, we get:
     \[
     (-1)^n - 1 \equiv 0 \pmod{17}
     \]
   - This implies:
     \[
     (-1)^n \equiv 1 \pmod{17}
     \]
   - For this to hold, \( n \) must be even because \( (-1)^n = 1 \) if and only if \( n \) is even.

Since \( n \) must be even for both conditions to hold, we conclude that \( N \) is divisible by \( 323 \) for all even integers \( n \).

The final answer is \( \boxed{ n } \) must be even.