Macedonian Mathematical Competitions

We have that
$$n^3 + 3n^2 - n - 3 = n^2(n+3) - (n+3) = (n+3)(n^2 - 1) = (n-1)(n+1)(n+3).$$
The numbers $n-1$, $n+1$, $n+3$ are even and one from $n-1$, $n+1$ is divisible by 4. Therefore the given number is divisible by 16. If $n$ is divisible by 3, then $n+3$ is divisible by 3. If $n$ is not divisible by 3, then the remainder is 1 or -1, from where we obtain that one of the numbers $n-1$ and $n+1$ is divisible by 3. Because of this, the given number is divisible by $3 \cdot 16 = 48$.

The number $n$ is odd, so $n = 2k-1$, for $k \in \mathbb{N}$.
$$(2k-1)^3 + 3(2k-1)^2 - (2k-1) - 3 = 8k^3 - 8k = 8k(k^2-1) = 8k(k-1)(k+1).$$
One of the numbers $k-1$, $k$, $k+1$ is divisible by 3 and at least one of them is even. Because of that $8k(k-1)(k+1)$ is divisible by $8 \cdot 3 \cdot 2 = 48$.