Question:
For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd?

Answer:
$\binom{n}{2} = \frac{n(n-1)}{2}$. In order for this fraction to be odd, neither $n$ nor $n-1$ can be divisible by $4$, because only one of $n$ and $n-1$ can be even. There are $25$ integers where $n$ is divisible by $4$, namely the multiples of $4$ from $4$ to $100$. There are $24$ integers where $n-1$ is divisible by $4$. We can obtain these integers by incrementing all the multiples of $4$ by $1$, but we must not include $100$ since $100+1 = 101 > 100$. Therefore, there are $49$ invalid integers, so there are $99 - 49 = \boxed{50}$ valid integers.