Positive real numbers are arranged in the form: 
$ 1 \ \ \ 3 \ \ \ 6 \ \ \ 10 \ \ \ 15 ...$
$ 2 \ \ \ 5 \ \ \ 9 \ \ \ 14 ...$
$ 4 \ \ \ 8 \ \ \ 13 ...$
$ 7 \ \ \ 12 ...$
$ 11 ...$
Find the number of the line and column where the number 2002 stays.
1. **Understanding the arrangement**: The given sequence of numbers is arranged in a specific pattern. To better understand the pattern, we can rotate the array 45 degrees clockwise. This gives us a triangular arrangement:
   \[
   \begin{array}{cccccc}
       &   &   & 1 &   &   \\
       &   & 2 &   & 3 &   \\
       & 4 &   & 5 &   & 6 \\
     7 &   & 8 &   & 9 & 10 \\
     \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
   \end{array}
   \]

2. **Identifying the pattern**: The first number of row \( n+1 \) is greater than the \( n \)-th triangular number by 1. The \( n \)-th triangular number \( T_n \) is given by:
   \[
   T_n = \frac{n(n+1)}{2}
   \]
   Therefore, the first number of row \( n+1 \) is:
   \[
   a_n = T_n + 1 = \frac{n(n+1)}{2} + 1
   \]

3. **Finding the row for 2002**: We need to find the largest \( n \) such that \( a_n \leq 2002 \). This means solving:
   \[
   \frac{n(n+1)}{2} + 1 \leq 2002
   \]
   Simplifying, we get:
   \[
   \frac{n(n+1)}{2} \leq 2001
   \]
   \[
   n(n+1) \leq 4002
   \]
   Solving the quadratic inequality:
   \[
   n^2 + n - 4002 \leq 0
   \]
   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 1 \), and \( c = -4002 \):
   \[
   n = \frac{-1 \pm \sqrt{1 + 4 \cdot 4002}}{2} = \frac{-1 \pm \sqrt{16009}}{2}
   \]
   Approximating \( \sqrt{16009} \approx 126.5 \):
   \[
   n \approx \frac{-1 + 126.5}{2} \approx 62.75
   \]
   Since \( n \) must be an integer, we take \( n = 62 \).

4. **Calculating the first number in the 63rd row**: 
   \[
   a_{62} = \frac{62 \cdot 63}{2} + 1 = 1954
   \]
   Since \( 2002 - 1954 = 48 \), 2002 is the 49th number in the 63rd row.

5. **Reverting the array back**: When we revert the array back to its original form, we see that the 63rd row corresponds to the 15th row in the original arrangement. Therefore, 2002 is in the 15th row and the 49th column.

The final answer is \(\boxed{(15, 49)}\)