e b $ contributes 4 lattice points (due to sign combinations), and symmetric pairs contribute similarly. But since $ a $ and $ b $ must both be odd (always true), and $ ab = 2025 $, we count all ordered pairs $ (a,b) $ with $ ab = 2025 $. There are 15 positive divisors, so 15 positive factor pairs $ (a,b) $, and 15 negative ones $ (-a,-b) $. Each gives integer $ x, y $. So total 30 pairs. Each pair yields a unique lattice point. Thus, there are $ oxed30 $ lattice points on the hyperbola. - Crosslake