Triangular Numbers are Simple

Let $i, j$ be positive integers such that $n + 2015 = \frac{i^{2} + i}{2}$ and $n - 2015 = \frac{j^{2} + j}{2}.$ Note that this implies that $i > j.$ If we subtract the second equation from the first, we get

$\begin{aligned} (n + 2015) - (n - 2015) & = \frac{i^{2} + i}{2} - \frac{j^{2} + j}{2} \\ 4030 & = \frac{i^{2} + i - j^{2} - j}{2} \\ 8060 & = i^{2} - j^{2} + i - j \\ 8060 & = (i + j)(i - j) + i - j \\ 8060 & = (i + j + 1)(i - j). \end{aligned}$

In order to minimize $n$ , we must let the "gap" between $i + j + 1$ and $i - j$ be as small as possible. Also, one of the expressions must be a multiple of $4$ , since they have different parities and $8060$ has a factor of $4$ . Finally, $i + j + 1 > i - j$ because $i$ and $j$ are positive. $i + j + 1 = 124$ and $i - j = 65$ is the setting we seek. Solving this gives $j = 29,$ so

$n = \frac{29^2 + 29}{2} + 2015 = 435 + 2015 = \boxed{2450}.$