Junior Turkish Mathematical Olympiad

Let $m^2 = n^2 + 2025n + 2010 \cdot 15$ where $m$ is a positive integer. Then $(2n + 2m + 2025)(2n - 2m + 2025) = 1995^2$. Hence $n = (A + B - 4050)/4$ and $m = (A - B)/4$

where $A$ and $B$ are integers such that $AB = 1995^2$, $A > B$ and $A + B > 4050$. Since $1995^2 \equiv 1 \pmod 4$, these formulas always give integer values for $m$ and $n$.
Let $f(x) = x + \frac{1995^2}{x}$. Then $f(x_2) - f(x_1) = (x_2 - x_1)\left(1 - \frac{1995^2}{x_1x_2}\right) > 0$ for $x_2 > x_1 > 1995$.
The first divisor of $1995^2 = 3^2 \cdot 5^2 \cdot 7^2 \cdot 19^2$ after $1995$ is $A = 2025 = 3^2 \cdot 5 \cdot 7^2$ for which $f(A) > 4050$ fails. The next one $A = 2527 = 7 \cdot 19^2$ and hence all the greater divisors satisfy $f(A) > 4050$. The answer is $((2+1)^4 - 1)/2 - 1 = 39$.