Does there exist a subset $E$ of the set $N$ of all positive integers such that none of the elements in $E$ can be presented as a sum of at least two other (not necessarily distinct) elements from $E$ ? 

(E. Barabanov)
To determine if there exists a subset \( E \) of the set \( \mathbb{N} \) of all positive integers such that none of the elements in \( E \) can be presented as a sum of at least two other (not necessarily distinct) elements from \( E \), we need to find a set \( E \) that satisfies this condition.

1. **Define the set \( E \):**
   Let's consider the set \( E = \{5, 14, 22\} \).

2. **Check if 5 can be represented as a sum of at least two elements from \( E \):**
   - The only possible sums involving 5 are \( 5 + 5 \), \( 5 + 14 \), and \( 5 + 22 \).
   - None of these sums equal 5, so 5 cannot be represented as a sum of at least two elements from \( E \).

3. **Check if 14 can be represented as a sum of at least two elements from \( E \):**
   - The only possible sums involving 14 are \( 14 + 5 \), \( 14 + 14 \), and \( 14 + 22 \).
   - None of these sums equal 14, so 14 cannot be represented as a sum of at least two elements from \( E \).

4. **Check if 22 can be represented as a sum of at least two elements from \( E \):**
   - The only possible sums involving 22 are \( 22 + 5 \), \( 22 + 14 \), and \( 22 + 22 \).
   - We need to check if 22 can be written as \( 5x + 14y \) where \( x \) and \( y \) are natural numbers.
   - Solving \( 5x + 14y = 22 \):
     \[
     5x + 14y = 22
     \]
     - If \( y = 1 \), then:
       \[
       5x + 14 = 22 \implies 5x = 8 \implies x = \frac{8}{5}
       \]
       - Since \( x \) must be a natural number, \( x = \frac{8}{5} \) is not possible.
     - If \( y = 0 \), then:
       \[
       5x = 22 \implies x = \frac{22}{5}
       \]
       - Since \( x \) must be a natural number, \( x = \frac{22}{5} \) is not possible.
     - Therefore, 22 cannot be represented as a sum of at least two elements from \( E \).

Since none of the elements in \( E = \{5, 14, 22\} \) can be represented as a sum of at least two other elements from \( E \), the set \( E \) satisfies the given condition.

The final answer is \( \boxed{ E = \{5, 14, 22\} } \)