Suppose that $m$ and $n$ are relatively prime positive integers with $A = \tfrac mn$, where
\[ A = \frac{2+4+6+\dots+2014}{1+3+5+\dots+2013} - \frac{1+3+5+\dots+2013}{2+4+6+\dots+2014}. \] Find $m$. In other words, find the numerator of $A$ when $A$ is written as a fraction in simplest form.

[i]Proposed by Evan Chen[/i]
1. First, we need to find the sum of the even numbers from 2 to 2014. This sequence is an arithmetic series with the first term \(a = 2\), the common difference \(d = 2\), and the last term \(l = 2014\). The number of terms \(n\) in this series can be found using the formula for the \(n\)-th term of an arithmetic sequence:
   \[
   l = a + (n-1)d \implies 2014 = 2 + (n-1) \cdot 2 \implies 2014 = 2n \implies n = 1007
   \]
   The sum \(S\) of the first \(n\) terms of an arithmetic series is given by:
   \[
   S = \frac{n}{2} (a + l) = \frac{1007}{2} (2 + 2014) = \frac{1007 \cdot 2016}{2} = 1007 \cdot 1008
   \]

2. Next, we need to find the sum of the odd numbers from 1 to 2013. This sequence is also an arithmetic series with the first term \(a = 1\), the common difference \(d = 2\), and the last term \(l = 2013\). The number of terms \(n\) in this series can be found similarly:
   \[
   l = a + (n-1)d \implies 2013 = 1 + (n-1) \cdot 2 \implies 2013 = 2n - 1 \implies 2014 = 2n \implies n = 1007
   \]
   The sum \(S\) of the first \(n\) terms of this arithmetic series is:
   \[
   S = \frac{n}{2} (a + l) = \frac{1007}{2} (1 + 2013) = \frac{1007 \cdot 2014}{2} = 1007^2
   \]

3. Now, we can express \(A\) as follows:
   \[
   A = \frac{2 + 4 + 6 + \dots + 2014}{1 + 3 + 5 + \dots + 2013} - \frac{1 + 3 + 5 + \dots + 2013}{2 + 4 + 6 + \dots + 2014}
   \]
   Substituting the sums we found:
   \[
   A = \frac{1007 \cdot 1008}{1007^2} - \frac{1007^2}{1007 \cdot 1008}
   \]

4. Simplify each term:
   \[
   \frac{1007 \cdot 1008}{1007^2} = \frac{1008}{1007}
   \]
   \[
   \frac{1007^2}{1007 \cdot 1008} = \frac{1007}{1008}
   \]

5. Combine the terms:
   \[
   A = \frac{1008}{1007} - \frac{1007}{1008}
   \]

6. Find a common denominator and simplify:
   \[
   A = \frac{1008^2 - 1007^2}{1007 \cdot 1008} = \frac{(1008 - 1007)(1008 + 1007)}{1007 \cdot 1008} = \frac{1 \cdot 2015}{1007 \cdot 1008}
   \]

7. Since 2015 is relatively prime to \(1007 \cdot 1008\), the fraction is already in its simplest form. Therefore, the numerator \(m\) is:
   \[
   \boxed{2015}
   \]