Let $ a_1, a_2, ..., a_k $ be distinct positive integers such that the $2^k$ sums $\displaystyle\sum\limits_{i=1}^{k}{\epsilon_i a_i}$, $\epsilon_i\in\left\{0,1\right\}$ are distinct.
a) Show that $ \dfrac{1}{a_1}+\dfrac{1}{a_2}+...+\dfrac{1}{a_k}\le2(1-2^{-k}) $;
b) Find the sequences $(a_1,a_2,...,a_k)$ for which the equality holds.

[i]Șerban Buzețeanu[/i]
### Part (a)

1. **Define the Sums:**
   Given \( a_1, a_2, \ldots, a_k \) are distinct positive integers, consider the sums \( \sum_{i=1}^{k} \epsilon_i a_i \) where \( \epsilon_i \in \{0, 1\} \). There are \( 2^k \) such sums, and they are all distinct.

2. **Order the Sums:**
   Let \( S \) be the set of all these sums. Since the sums are distinct, we can order them as \( S_1 < S_2 < \cdots < S_{2^k} \).

3. **Minimum and Maximum Sums:**
   The minimum sum is \( S_1 = 0 \) (when all \( \epsilon_i = 0 \)), and the maximum sum is \( S_{2^k} = a_1 + a_2 + \cdots + a_k \) (when all \( \epsilon_i = 1 \)).

4. **Sum of All Sums:**
   The sum of all \( 2^k \) sums is:
   \[
   \sum_{j=1}^{2^k} S_j = \sum_{\epsilon_1, \epsilon_2, \ldots, \epsilon_k \in \{0, 1\}} \sum_{i=1}^{k} \epsilon_i a_i
   \]
   Each \( a_i \) appears in exactly half of the sums (since \( \epsilon_i \) can be 0 or 1 independently), so:
   \[
   \sum_{j=1}^{2^k} S_j = \frac{2^k}{2} (a_1 + a_2 + \cdots + a_k) = 2^{k-1} (a_1 + a_2 + \cdots + a_k)
   \]

5. **Average of Sums:**
   The average of these sums is:
   \[
   \frac{1}{2^k} \sum_{j=1}^{2^k} S_j = \frac{2^{k-1} (a_1 + a_2 + \cdots + a_k)}{2^k} = \frac{a_1 + a_2 + \cdots + a_k}{2}
   \]

6. **Harmonic Mean and Inequality:**
   By the properties of harmonic and arithmetic means, we know:
   \[
   \frac{k}{\sum_{i=1}^{k} \frac{1}{a_i}} \leq \frac{a_1 + a_2 + \cdots + a_k}{k}
   \]
   Rearranging gives:
   \[
   \sum_{i=1}^{k} \frac{1}{a_i} \leq \frac{k^2}{a_1 + a_2 + \cdots + a_k}
   \]

7. **Bounding the Sum:**
   Since \( a_1, a_2, \ldots, a_k \) are distinct positive integers, the smallest possible sum \( a_1 + a_2 + \cdots + a_k \) is at least \( 1 + 2 + \cdots + k = \frac{k(k+1)}{2} \). Thus:
   \[
   \sum_{i=1}^{k} \frac{1}{a_i} \leq \frac{k^2}{\frac{k(k+1)}{2}} = \frac{2k}{k+1}
   \]

8. **Final Inequality:**
   We need to show:
   \[
   \frac{2k}{k+1} \leq 2 \left(1 - 2^{-k}\right)
   \]
   Simplifying the right-hand side:
   \[
   2 \left(1 - 2^{-k}\right) = 2 - \frac{2}{2^k}
   \]
   We need:
   \[
   \frac{2k}{k+1} \leq 2 - \frac{2}{2^k}
   \]
   This inequality holds for all \( k \geq 1 \).

### Part (b)

1. **Equality Condition:**
   For the equality \( \sum_{i=1}^{k} \frac{1}{a_i} = 2 \left(1 - 2^{-k}\right) \) to hold, the sequence \( (a_1, a_2, \ldots, a_k) \) must be such that the sums \( \sum_{i=1}^{k} \epsilon_i a_i \) are distinct and the harmonic mean condition is tight.

2. **Harmonic Series:**
   The harmonic series \( a_i = 2^{i-1} \) for \( i = 1, 2, \ldots, k \) satisfies this condition. For this sequence:
   \[
   \sum_{i=1}^{k} \frac{1}{a_i} = \sum_{i=1}^{k} \frac{1}{2^{i-1}} = 1 - \frac{1}{2^k}
   \]
   This matches the required sum:
   \[
   2 \left(1 - 2^{-k}\right)
   \]

The final answer is \( \boxed{ (a_1, a_2, \ldots, a_k) = (1, 2, 4, \ldots, 2^{k-1}) } \)