5. For what positive integer $n$ is $20^{n}+16^{n}-3^{n}-1$ divisible by 323?
5. Let $a \mid b$ denote $a$ divides $b$. Since for polynomials $a-b \mid a^{k}-b^{k}, a+b \mid a^{2 k-1}+b^{2 k-1}, k \in N$, we have:
(1) When $2 \mid n$, $17=20-3 \mid 20^{n}-3^{n}$,
$17 \mid 255=256-1 \mid 256^{\frac{n}{2}}-1=16^{n}-1$.
Thus, $17 \mid A$;
Similarly, $19 \mid 20^n - 1$,
$19 \mid 247 \mid 256^{\frac{n}{2}}-9^{\frac{n}{2}}=16^{n}-3^{n}, 19 \mid A$.
Therefore, $17 \cdot 19=323 \mid A$.
(2) When $2 \nmid n$, $17 \mid 20^{n}-3^{n}$.
$$
17=16+1 \mid 16^{n}+1 \text {, }
$$

Thus, $17 \nmid 16^{n}-1, 17 \nmid A, 323 \nmid A$.
In conclusion, the $n$ that satisfies the problem is any positive even number.