We are given the recurrence relation $t_n = \frac{n-3}{n-1} t_{n-2}$ for $n \ge 3$, along with the initial terms $t_1 = 1$ and $t_2 = -1$.
We need to calculate the sum:
\[ S = \frac{1}{t_2} + \frac{1}{t_4} + \frac{1}{t_6} + \dots + \frac{1}{t_{2022}} + \frac{1}{t_{2024}} \]
We first observe the pattern in the terms generated by the recurrence relation. Using the recurrence, we can calculate the first few terms:
\(t_3\)\(= \frac{3-3}{3-1} t_1 = 0\)
\(t_4 = \frac{4-3}{4-1} t_2 = \frac{1}{3} \times (-1) = -\frac{1}{3}\)
\(t_5 = \frac{5-3}{5-1} t_3 = \frac{2}{4} \times 0 = 0\)
\(t_6 = \frac{6-3}{6-1} t_4 = \frac{3}{5}\)\(\times \left(-\frac{1}{3}\right) = -\frac{1}{5}\)
From this, we notice that $t_n$ for even $n$ follows the pattern:
\[ t_2 = -1, \quad t_4 = -\frac{1}{3}, \quad t_6 = -\frac{1}{5}, \dots \]
Thus, the values of $t_n$ for even $n$ are the negative reciprocals of the odd numbers starting from 1, i.e., $t_n = -\frac{1}{n-1}$.
Now, the sum is:
\[ S = \sum_{k=1}^{1012} \frac{1}{t_{2k}} = \sum_{k=1}^{1012} -(2k-1) = -\sum_{k=1}^{1012} (2k-1) \]
The sum of the first 1012 odd numbers is $1012^2$, so:
\[ S = -1012^2 = -1024144 \]
Thus, the correct answer is Option (1).