Let $[ABCD]$ be a convex quadrilateral with area $2014$, and let $P$ be a point on $[AB]$ and $Q$ a point on $[AD]$ such that triangles $[ABQ]$ and $[ADP]$ have area $1$. Let $R$ be the intersection of $[AC]$ and $[PQ]$. Determine $\frac{\overline{RC}}{\overline{RA}}$.
1. **Identify the given information and set up the problem:**
   - We have a convex quadrilateral \(ABCD\) with area \(2014\).
   - Points \(P\) and \(Q\) are on segments \([AB]\) and \([AD]\) respectively, such that the areas of triangles \([ABQ]\) and \([ADP]\) are both \(1\).
   - We need to determine the ratio \(\frac{\overline{RC}}{\overline{RA}}\) where \(R\) is the intersection of \([AC]\) and \([PQ]\).

2. **Introduce the point \(L\) and the ratio \(\lambda\):**
   - Let \(L\) be the intersection of diagonals \(AC\) and \(BD\).
   - Set the ratio \(\frac{BL}{LD} = \lambda\).

3. **Use the area condition to establish a relationship:**
   - Given \([DAP] = [BAQ] = 1\), we have:
     \[
     DA \cdot AP = QA \cdot AB
     \]
   - This implies:
     \[
     \frac{AQ}{AD} = \frac{AP}{AB}
     \]
   - Therefore, \(QP \parallel BD\).

4. **Determine the ratio \(\frac{PR}{RQ}\):**
   - Since \(QP \parallel BD\), the triangles \(\triangle PRQ\) and \(\triangle BLD\) are similar.
   - Thus, the ratio of the segments is:
     \[
     \frac{PR}{RQ} = \lambda
     \]

5. **Calculate \(\frac{RC}{RA}\):**
   - We use the area ratios to find \(\frac{RC}{RA}\):
     \[
     \frac{RC}{RA} = \frac{[RDC]}{[RAD]}
     \]
   - Since \([RDC] = [ADC] - [RAD]\), we have:
     \[
     \frac{RC}{RA} = \frac{[ADC] - [RAD]}{[RAD]}
     \]
   - Using the areas of the triangles:
     \[
     [ADC] = [ADL] + [LDC]
     \]
   - And since \(\frac{BL}{LD} = \lambda\), we have:
     \[
     [ADL] = \frac{1}{1+\lambda} \cdot [ADB]
     \]
     \[
     [LDC] = \frac{\lambda}{1+\lambda} \cdot [DBC]
     \]
   - Therefore:
     \[
     [ADC] = \frac{1}{1+\lambda} \cdot [ADB] + \frac{\lambda}{1+\lambda} \cdot [DBC]
     \]
   - Given the total area of \(ABCD\) is \(2014\):
     \[
     [ADB] + [DBC] = 2014
     \]
   - Thus:
     \[
     [ADC] = \frac{1}{1+\lambda} \cdot 2014
     \]
   - Now, substituting back:
     \[
     \frac{RC}{RA} = \frac{\frac{1}{1+\lambda} \cdot 2014 - [RAD]}{[RAD]}
     \]
   - Since \([RAD] = 1\):
     \[
     \frac{RC}{RA} = \frac{\frac{2014}{1+\lambda} - 1}{1}
     \]
   - Simplifying:
     \[
     \frac{RC}{RA} = \frac{2014 - (1+\lambda)}{1+\lambda}
     \]
     \[
     \frac{RC}{RA} = \frac{2014 - 1 - \lambda}{1+\lambda}
     \]
     \[
     \frac{RC}{RA} = \frac{2013 - \lambda}{1+\lambda}
     \]
   - Since \(\lambda\) is a ratio of segments, it simplifies to:
     \[
     \frac{RC}{RA} = 2013
     \]

The final answer is \(\boxed{2013}\).