The sum of the first \(n\) integers can be written as:
$$ 1 + 2 + 3 + ... + (n-1) + n = \frac{(n)(n+1)}{2} $$
This means:
$$\begin{align} 1 + 2 + 3 + ... + (n-1) &= \frac{(n)(n+1)}{2} - n \\ &= \frac{(n)(n-1)}{2} \end{align}$$
Either \(n\) is even or \(n-1\) is even. If \(\frac{n-1}{2}\) is an integer, then \([(n)\frac{(n-1)}{2} ≡ 0 \bmod n]\).
If \((n/2)\) is an integer, then \([(n-1)(n/2) ≡ -(n/2) \bmod n]\), as \([(n-1) ≡ -1 \bmod n]\). Since \((n/2)\) and \(-(n/2)\) have a difference of \(n\), then \([(n-1)(n/2) ≡ (n/2) \bmod n]\).
This means if \(n\) is a positive integer, then \([\frac{(n)(n-1)}{2} ≡ 0 \bmod n]\) if and only if \(n\) is odd, otherwise \([\frac{(n)(n-1)}{2} ≡ \frac{n}{2} \bmod n]\).