Solution:

We shall observe straight lines as on the next picture. We can call these lines diagonals.

On the first diagonal is number $1$.
On the second diagonal are two numbers: $2$ and $3$.
On the third diagonal are three numbers: $4$, $5$ and $6$.
On the $n$-th diagonal are $n$ numbers. These numbers are greater than $\frac{(n-1)n}{2}$ and not greater than $\frac{n(n+1)}{2}$.
On the first $n$ diagonals are $1+2+3+\ldots+n=\frac{n(n+1)}{2}$ numbers.
If $m$ is in the $k$-th row, $l$-th column and on the $n$-th diagonal, then it is $m=\frac{(n-1)n}{2}+l$ and $n+1=k+l$. So, $m=\frac{(k+l-2)(k+l-1)}{2}+l$.
We have to find such numbers $n$, $k$ and $l$ for which:
$$\begin{gathered} \frac{(n-1)n}{2}<2015 \leq \frac{n(n+1)}{2} \\ n+1=k+l \\ 2015=\frac{(k+l-2)(k+l-1)}{2}+l \end{gathered}$$
From (1), (2), (3):
$$n^2-n<4030 \leq n^2+n \implies n=63,\ k+l=64,\ 2015=\frac{(64-2)(64-1)}{2}+l$$
So $2015=2016+l$ so $l=2015-2016=-1$ (but this is not possible, so check calculation).
Actually, $\frac{(64-2)(64-1)}{2}=2016$, so $l=2015-2016=-1$.
But $\frac{(63)(62)}{2}=1953$, $\frac{64\cdot 63}{2}=2016$.
So $2015=1953+l$, $l=2015-1953=62$.
So $k+l=64$, $l=62$, so $k=2$.

Therefore, $2015$ is located in the second row and $62$-th column.

Firstly, we can see that the first elements of the columns are triangular numbers. If $a_{i}$ is the first element of the line $i$, we have $a_{i}=\frac{i(i-1)}{2}$.
The second element of the first row is obtained by adding to the first element $2$.
The second element of the second row is obtained by adding to the first element $3$.
And so on, then the second element on the $n$-th row is obtained by adding to the first element $n+1$.
Then the third element of the $n$-th row is obtained by adding $n+2$, and the $k$-th element of it is obtained by adding $k$.
Since the first element of the $n$-th row is $\frac{(n-1)n}{2}+1$, the second one is $\frac{(n-1)n}{2}+1+(n+1)=\frac{n(n+1)}{2}+2$.
The third one $\frac{n(n+1)}{2}+1+(n+2)=\frac{(n+1)(n+2)}{2}+3$, so the $k$-th one should be $\frac{(n+k-2)(n+k-1)}{2}+k$.

$$\frac{(n+k-2)(n+k-1)}{2}+k=2015 \Leftrightarrow n^2+n(2k-3)+k^2-k-4028=0$$
To have a positive integer solution, $(2k-3)^2-4\left(k^2-k-4028\right)=16121-8k$ must be a perfect square.
From $16121-8k=x^2$, it is noticed that the maximum of $x$ is $126$ (since $k>0$).
Simultaneously, $x$ is odd, so $x\leq 125$.
$16121-8k=x^2 \Leftrightarrow 125^2+496-8k=x^2$
So $496-8k=0$, from that $k=62$.
From that we can find $n=2$.
So $2015$ is located on the second row and $62$-th column.