Solution:

The closest perfect square to $2015$ is $2025 = 45^2$ which means that only the rightmost side will be incomplete while the required diagonal would still have a total of $45$ entries.

Looking at the values on the diagonal, we see that the numbers on the diagonal above $1$ have a common second difference. This suggests that this sequence satisfies a quadratic function of the form $f(n) = a n^2 + b n + c$. Since $f(1) = 1$, $f(2) = 3$, $f(3) = 13$, solving a simple system of three equations gives us $f(n) = 4 n^2 - 10 n + 7$, $1 \leq n \leq 23$. On the other hand, the numbers on the diagonal below $1$ also have a common second difference. This gives a sequence $g(n) = d n^2 + e n + f$ with $g(1) = 1$, $g(2) = 7$, and $g(3) = 21$. By solving a similar system as above, we obtain $g(n) = 4 n^2 - 6 n + 3$, where $1 \leq n \leq 23$. Taking the sum of these two sequences of numbers, we have

$$\begin{aligned} \sum_{n=1}^{23}[f(n)+g(n)] & = \sum_{n=1}^{23}\left(8 n^2 - 16 n + 10\right) \\ & = 8 \sum_{n=1}^{23} n^2 - 16 \sum_{n=1}^{23} n + \sum_{n=1}^{23} 10 \\ & = 8\left[\frac{(23)(24)(47)}{6}\right] - 16\left[\frac{(23)(24)}{2}\right] + 10(23) \\ & = 30,406 \end{aligned}$$

Since $1$ is counted twice, the required sum must be $30,406 - 1 = 30,405$.

Solution:

Filling the square with a few more numbers enables us to see that the boxed numbers
$$1, 3, 7, 13, 21, 31, \ldots, 1981$$
satisfy the recurrence relation $a_{1} = 1$ and $(\forall n \in \mathbb{N})\ a_{n+1} = a_{n} + 2n$. The associated homogeneous recurrence relation is solved by $a_{n}^{(h)} \equiv 1$. Testing a particular solution of the form $a_{n}^{(p)} = n(c n + d)$, we see that $c = 1$ and $d = -1$. Therefore, the solution to the nonhomogeneous recurrence relation is $a_{n} = n^2 - n + 1$. The last boxed number $1981$ corresponds to $n = 45$. Therefore,
$$\sum_{n=1}^{45} \left(n^2 - n + 1\right) = \frac{45 \cdot 46 \cdot 91}{6} - \frac{45 \cdot 46}{2} + 45 = 30,405$$