Find all positive reals $x,y,z $ such that \[2x-2y+\dfrac1z = \dfrac1{2014},\hspace{0.5em} 2y-2z +\dfrac1x = \dfrac1{2014},\hspace{0.5em}\text{and}\hspace{0.5em} 2z-2x+ \dfrac1y = \dfrac1{2014}.\]
To find all positive real numbers \(x, y, z\) that satisfy the given system of equations:
\[
2x - 2y + \frac{1}{z} = \frac{1}{2014},
\]
\[
2y - 2z + \frac{1}{x} = \frac{1}{2014},
\]
\[
2z - 2x + \frac{1}{y} = \frac{1}{2014},
\]
we can follow these steps:

1. **Add all three equations:**
   \[
   (2x - 2y + \frac{1}{z}) + (2y - 2z + \frac{1}{x}) + (2z - 2x + \frac{1}{y}) = \frac{1}{2014} + \frac{1}{2014} + \frac{1}{2014}.
   \]
   Simplifying the left-hand side:
   \[
   2x - 2y + 2y - 2z + 2z - 2x + \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = \frac{3}{2014}.
   \]
   This reduces to:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2014}.
   \]

2. **Express each equation in terms of \(x, y, z\):**
   \[
   2x - 2y + \frac{1}{z} = \frac{1}{2014},
   \]
   \[
   2y - 2z + \frac{1}{x} = \frac{1}{2014},
   \]
   \[
   2z - 2x + \frac{1}{y} = \frac{1}{2014}.
   \]

3. **Substitute \(x = 2014a\), \(y = 2014b\), \(z = 2014c\):**
   \[
   2(2014a) - 2(2014b) + \frac{1}{2014c} = \frac{1}{2014},
   \]
   \[
   2(2014b) - 2(2014c) + \frac{1}{2014a} = \frac{1}{2014},
   \]
   \[
   2(2014c) - 2(2014a) + \frac{1}{2014b} = \frac{1}{2014}.
   \]
   Simplifying these equations:
   \[
   4028a - 4028b + \frac{1}{2014c} = \frac{1}{2014},
   \]
   \[
   4028b - 4028c + \frac{1}{2014a} = \frac{1}{2014},
   \]
   \[
   4028c - 4028a + \frac{1}{2014b} = \frac{1}{2014}.
   \]

4. **Simplify further:**
   \[
   4028(a - b) + \frac{1}{2014c} = \frac{1}{2014},
   \]
   \[
   4028(b - c) + \frac{1}{2014a} = \frac{1}{2014},
   \]
   \[
   4028(c - a) + \frac{1}{2014b} = \frac{1}{2014}.
   \]

5. **Multiply through by 2014:**
   \[
   4028 \cdot 2014(a - b) + \frac{1}{c} = 1,
   \]
   \[
   4028 \cdot 2014(b - c) + \frac{1}{a} = 1,
   \]
   \[
   4028 \cdot 2014(c - a) + \frac{1}{b} = 1.
   \]

6. **Simplify the constants:**
   \[
   8103952(a - b) + \frac{1}{c} = 1,
   \]
   \[
   8103952(b - c) + \frac{1}{a} = 1,
   \]
   \[
   8103952(c - a) + \frac{1}{b} = 1.
   \]

7. **Assume \(a = b = c\):**
   If \(a = b = c\), then:
   \[
   8103952(a - a) + \frac{1}{a} = 1,
   \]
   \[
   \frac{1}{a} = 1.
   \]
   Thus, \(a = 1\).

8. **Substitute back:**
   \[
   x = 2014a = 2014,
   \]
   \[
   y = 2014b = 2014,
   \]
   \[
   z = 2014c = 2014.
   \]

Therefore, the only positive real solution is \(x = y = z = 2014\).

The final answer is \(\boxed{(2014, 2014, 2014)}\).