We will analyze the pattern in the table to determine where the number $2015$ is located. The arrangement of numbers in the table has a specific formulation in terms of line and column indices. We will first observe the pattern, derive the general formula for numbers at position $(i, j)$, and then use it to find the location of the number $2015$.

1. Pattern Observation:

Looking at the first few values:
- First row: $1, 3, 6, 10, 15, \ldots$ (These are triangular numbers: $T_n = \frac{n(n+1)}{2}$)
- First column: $1, 2, 4, 7, 11, 16, \ldots$ (These numbers follow a pattern where the difference between terms increases by 1: $(2-1), (4-2), (7-4), \ldots$)

2. Row Analysis:

The $j$-th number in the first row is the $j$-th triangular number:
$$T_j = \frac{j(j+1)}{2}.$$

3. Column Analysis:

The $i$-th number in the first column is the sum of the first $i-1$ natural numbers plus 1:
$$S_i = 1 + \sum_{k=1}^{i-1} k = 1 + \frac{(i-1)i}{2}.$$

4. General Formula:

For a number in the table located at position $(i, j)$, the formula is:
$$a_{i,j} = S_i + T_{j-1}.$$
Simplified, this is:
$$a_{i,j} = 1 + \frac{(i-1)i}{2} + \frac{(j-1)j}{2}.$$

5. **Solving for $2015$**:

We need to find $i$ and $j$ such that:
$$1 + \frac{(i-1)i}{2} + \frac{(j-1)j}{2} = 2015.$$

6. Approximations:

Start by trying values for $i$:
- For $i = 62$:
$$S_{62} = 1 + \frac{(62-1) \cdot 62}{2} = 1 + \frac{61 \cdot 62}{2} = 1892.$$
- Now solve:
$$1892 + T_{j-1} = 2015 \Rightarrow T_{j-1} = 2015 - 1892 = 123.$$
- Find $j$:
$$\frac{(j-1)j}{2} = 123 \quad \Rightarrow \quad (j-1)j = 246.$$
- Solving $j^2 - j - 246 = 0$ using the quadratic formula yields:
$$j = \frac{1 + \sqrt{1 + 4 \cdot 246}}{2} = \frac{1 + \sqrt{985}}{2} = \frac{1 + 31}{2} = 16$$
However, this solution for $(j-1) = 15$; thus, $j = 2$.

7. Conclusion:

The number $2015$ is located in line 62 and column 2 of the table. Therefore, the answer is:

$$\boxed{(62, 2)}$$

This approach successfully determines the correct position of the number $2015$ in the table.