Note that $(n + 15)(n + 2010)$ is a perfect square iff $4(n + 15)(n + 2010)$ is a perfect square, or $(2n + 30)(2n + 4020)$ is a perfect square. Let $2n + 2025 = a$, and the perfect square be $b^2$; then $(a - 1995)(a + 1995) = b^2$. Hence, $(a + b)(a - b) = 1995^2$, or $3^2 \cdot 5^2 \cdot 7^2 \cdot 19^2$. This has $81$ factors :O Note that $a + b = a - b = 1995$ cannot be a solution, otherwise $b = 0$. Also, $a > 2025$ beacuse $2n + 2025 = a$. The only pair that does not satisfy this is $(1805,2205)$. Finally, $a + b > a - b$. So we take $81 - 3$, or $78$, and divide by $2$ to get $\boxed{39}$ solutions.

Note that $(n + 15)(n + 2010)$ is a perfect square iff $4(n + 15)(n + 2010)$ is a perfect square, or $(2n + 30)(2n + 4020)$ is a perfect square. Let $2n + 2025 = a$, and the perfect square be $b^2$; then $(a - 1995)(a + 1995) = b^2$. Hence, $(a + b)(a - b) = 1995^2$, or $3^2 \cdot 5^2 \cdot 7^2 \cdot 19^2$. This has $81$ factors :O Note that $a + b = a - b = 1995$ cannot be a solution, otherwise $b = 0$. Also, $a > 2025$ beacuse $2n + 2025 = a$. The only pair that does not satisfy this is $(1805,2205)$. Finally, $a + b > a - b$. So we take $81 - 3$, or $78$, and divide by $2$ to get $\boxed{39}$ solutions.