Near-Squared

We require:

$\Large n^2 + 2767n = k^2 \Rightarrow n = \frac{-2767 \pm \sqrt{2767^2 - 4(1)(-k^2)}}{2} = \frac{-2767 \pm \sqrt{4k^2 + 2767^2}}{2}$ (i)

In order for $n$ to be an integer, we require the discriminant in (i) to be a perfect square. That is, $4k^2 + 2767^2 = N^2 \Rightarrow 2767^2 = N^2-4k^2 = (N+2k)(N-2k)$ . The number $2767^2$ has divisors $1, 2767, 2767^2$ (Note: $2767$ is a prime), which gives us the following ordered-pairs:

$N+2k = 2767, N-2k = 2767 \Rightarrow N = 2767, k = 0$ (ii);

$N+2k = 2767^2, N-2k = 1 \Rightarrow N = 3828145, k =1914072$ (iii).

Substitution of the above values for $k$ in (ii) and (iii) back into the discriminant of (i) yields:

$n = 0, -2767, -1915456, 1912689$

of which $\boxed{n= 1912689}$ is the only positive integer solution.