Solution:

We are given:
t1 = 1, t2 = -1, and tn = ((n-3)/(n-1)) * t(n-2) for n ≥ 3.

Step 1: Compute first few terms.
t2 = -1
t4 = (4-3)/(4-1) × t2 = (1/3) × (-1) = -1/3
t6 = (6-3)/(6-1) × t4 = (3/5) × (-1/3) = -1/5
t8 = (8-3)/(8-1) × t6 = (5/7) × (-1/5) = -1/7

So, we see the pattern:
t2 = -1,  t4 = -1/3,  t6 = -1/5,  t8 = -1/7, ...

Step 2: General pattern.
t(2k) = -1/(2k-1)

Step 3: Inverse values.
1/t2 = -1
1/t4 = -3
1/t6 = -5
1/t8 = -7
...
1/t2024 = -2023

Step 4: Sum of series.
1/t2 + 1/t4 + 1/t6 + ... + 1/t2024
= -(1 + 3 + 5 + 7 + ... + 2023)

The series inside is the sum of first 1012 odd numbers.

Sum of first n odd numbers = n²
So, sum = (1012)² = 1024144

Final Answer:
= -1024144