Let $A=\{1,2,4,5,7,8,\dots\}$ the set with naturals not divisible by three. Find all values of $n$ such that exist $2n$ consecutive elements of $A$ which sum it´s $300$.
To solve the problem, we need to find all values of \( n \) such that there exist \( 2n \) consecutive elements of the set \( A \) whose sum is 300. The set \( A \) consists of natural numbers that are not divisible by 3.

### Case 1: First number is \( 3k+1 \)
1. The \( 2n \) consecutive numbers starting from \( 3k+1 \) are:
   \[
   3k+1, 3k+2, 3k+4, 3k+5, \ldots, 3k+3(n-1)+1, 3k+3(n-1)+2
   \]
2. The sum of these \( 2n \) numbers can be expressed as:
   \[
   \sum_{i=0}^{2n-1} (3k + 1 + \lfloor \frac{i}{2} \rfloor + i \mod 2)
   \]
   Simplifying, we get:
   \[
   \sum_{i=0}^{2n-1} (3k + 1 + \lfloor \frac{i}{2} \rfloor + i \mod 2) = 6kn + 3n^2
   \]
3. We set up the equation:
   \[
   6kn + 3n^2 = 300
   \]
   Dividing by 3:
   \[
   2kn + n^2 = 100
   \]
4. Factoring, we get:
   \[
   n(2k + n) = 100
   \]
5. We need to find integer solutions for \( n \) and \( k \). Let's try \( n = 2m \):
   \[
   2m(2k + 2m) = 100 \implies m(k + m) = 25
   \]
6. Possible pairs \((m, k)\) are:
   \[
   (1, 24), (5, 0)
   \]
   - For \( m = 1 \), \( n = 2 \), \( k = 24 \):
     \[
     n = 2, k = 24 \implies 2(2 \cdot 24 + 2) = 100
     \]
     The numbers are \( 73, 74, 76, 77 \).

### Case 2: First number is \( 3k+2 \)
1. The \( 2n \) consecutive numbers starting from \( 3k+2 \) are:
   \[
   3k+2, 3k+4, 3k+5, \ldots, 3k+3(n-1)+1, 3k+3(n-1)+2, 3k+3n+1
   \]
2. The sum of these \( 2n \) numbers can be expressed as:
   \[
   \sum_{i=0}^{2n-1} (3k + 2 + \lfloor \frac{i}{2} \rfloor + i \mod 2)
   \]
   Simplifying, we get:
   \[
   \sum_{i=0}^{2n-1} (3k + 2 + \lfloor \frac{i}{2} \rfloor + i \mod 2) = 6kn + 3n^2 + 3n
   \]
3. We set up the equation:
   \[
   6kn + 3n^2 + 3n = 300
   \]
   Dividing by 3:
   \[
   2kn + n^2 + n = 100
   \]
4. Factoring, we get:
   \[
   n(2k + n + 1) = 100
   \]
5. We need to find integer solutions for \( n \) and \( k \). Let's try \( n = 4 \):
   \[
   4(2k + 4 + 1) = 100 \implies 2k + 5 = 25 \implies k = 10
   \]
   - For \( n = 4 \), \( k = 10 \):
     \[
     n = 4, k = 10 \implies 4(2 \cdot 10 + 4 + 1) = 100
     \]
     The numbers are \( 32, 34, 35, \ldots, 43 \).

The final answer is \( \boxed{ n = 2 } \) and \( n = 4 \).