Determine all positive integers $a$ for which there exist exactly $2014$ positive integers $b$ such that $\displaystyle2\leq\frac{a}{b}\leq5$.
1. We start with the given inequality:
   \[
   2 \leq \frac{a}{b} \leq 5
   \]
   This can be rewritten as:
   \[
   2b \leq a \quad \text{and} \quad 5b \geq a
   \]
   Combining these inequalities, we get:
   \[
   \frac{a}{5} \leq b \leq \frac{a}{2}
   \]

2. We need to find the number of integers \( b \) that satisfy this inequality. The number of integers \( b \) in the interval \(\left[\frac{a}{5}, \frac{a}{2}\right]\) is given by:
   \[
   \left\lfloor \frac{a}{2} \right\rfloor - \left\lceil \frac{a}{5} \right\rceil + 1
   \]
   We are given that this number is exactly 2014:
   \[
   \left\lfloor \frac{a}{2} \right\rfloor - \left\lceil \frac{a}{5} \right\rfloor + 1 = 2014
   \]

3. To simplify, we approximate:
   \[
   \frac{a}{2} - \frac{a}{5} \approx 2014
   \]
   Solving for \( a \):
   \[
   \frac{a}{2} - \frac{a}{5} = 2014
   \]
   \[
   \frac{5a - 2a}{10} = 2014
   \]
   \[
   \frac{3a}{10} = 2014
   \]
   \[
   3a = 20140
   \]
   \[
   a = \frac{20140}{3} \approx 6713.33
   \]
   Since \( a \) must be an integer, we check the integers around 6713.

4. Checking \( a = 6710 \):
   \[
   \left\lfloor \frac{6710}{2} \right\rfloor - \left\lceil \frac{6710}{5} \right\rceil + 1 = 3355 - 1342 + 1 = 2014
   \]
   This works.

5. Checking \( a = 6712 \):
   \[
   \left\lfloor \frac{6712}{2} \right\rfloor - \left\lceil \frac{6712}{5} \right\rceil + 1 = 3356 - 1343 + 1 = 2014
   \]
   This works.

6. Checking \( a = 6713 \):
   \[
   \left\lfloor \frac{6713}{2} \right\rfloor - \left\lceil \frac{6713}{5} \right\rceil + 1 = 3356 - 1343 + 1 = 2014
   \]
   This works.

Therefore, the positive integers \( a \) that satisfy the condition are \( 6710, 6712, \) and \( 6713 \).

The final answer is \( \boxed{6710, 6712, 6713} \)