Consider two circles of radius one, and let $O$ and $O'$ denote their centers.  Point $M$ is selected on either circle.  If $OO' = 2014$, what is the largest possible area of triangle $OMO'$?

[i]Proposed by Evan Chen[/i]
1. **Positioning Point \( M \)**: Without loss of generality (WLOG), let point \( M \) be on the circle centered at \( O \). This simplifies our calculations as we can focus on the geometry involving \( O \), \( O' \), and \( M \).

2. **Area of Triangle \( OMO' \)**: The area of triangle \( OMO' \) can be calculated using the formula for the area of a triangle given two sides and the included angle:
   \[
   \text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\theta)
   \]
   Here, \(\text{side}_1 = OM = 1\) (radius of the circle centered at \( O \)), \(\text{side}_2 = OO' = 2014\), and \(\theta = \angle MOO'\).

3. **Maximizing the Area**: To maximize the area, we need to maximize \(\sin(\theta)\). The sine function reaches its maximum value of 1 when \(\theta = 90^\circ\). Therefore, the maximum area occurs when \(\angle MOO' = 90^\circ\).

4. **Calculating the Maximum Area**: Substituting the maximum value of \(\sin(\theta)\) into the area formula:
   \[
   \text{Area} = \frac{1}{2} \times 1 \times 2014 \times 1 = \frac{2014}{2} = 1007
   \]

Conclusion:
\[
\boxed{1007}
\]