Do there exist $ 14$ positive integers, upon increasing each of them by $ 1$,their product increases exactly $ 2008$ times?
1. We need to find 14 positive integers \( x_1, x_2, \ldots, x_{14} \) such that the product of each increased by 1 is exactly 2008 times the product of the original integers. Mathematically, this can be expressed as:
   \[
   \prod_{i=1}^{14} \frac{x_i + 1}{x_i} = 2008
   \]

2. We start by factoring 2008:
   \[
   2008 = 2^3 \times 251
   \]

3. We assume \( x_1 \ge x_2 \ge \cdots \ge x_{14} \). To simplify, we can choose \( x_1 \) such that \( \frac{x_1 + 1}{x_1} = 251 \). This gives:
   \[
   x_1 + 1 = 251x_1 \implies 250x_1 = 1 \implies x_1 = 250
   \]

4. Now, we need to satisfy:
   \[
   \prod_{i=2}^{14} \frac{x_i + 1}{x_i} = 2^3 = 8
   \]

5. We can choose \( x_2, x_3, x_4 \) such that \( \frac{x_i + 1}{x_i} = 2 \). This gives:
   \[
   x_i + 1 = 2x_i \implies x_i = 1 \quad \text{for} \quad i = 2, 3, 4
   \]

6. Now, we need to satisfy:
   \[
   \prod_{i=5}^{14} \frac{x_i + 1}{x_i} = 2^3 = 8
   \]

7. We can choose \( x_5, x_6, \ldots, x_{14} \) such that \( \frac{x_i + 1}{x_i} = 2 \). This gives:
   \[
   x_i + 1 = 2x_i \implies x_i = 1 \quad \text{for} \quad i = 5, 6, \ldots, 14
   \]

8. Therefore, the integers \( x_1, x_2, \ldots, x_{14} \) are:
   \[
   x_1 = 250, \quad x_2 = x_3 = x_4 = 1, \quad x_5 = x_6 = \cdots = x_{14} = 1
   \]

Conclusion:
\[
\boxed{x_1 = 250, \quad x_2 = x_3 = x_4 = 1, \quad x_5 = x_6 = \cdots = x_{14} = 1}
\]