Given is a positive integer $n \geq 2$. Determine the maximum value of the sum of natural numbers $k_1,k_2,...,k_n$ satisfying the condition:
$k_1^3+k_2^3+ \dots +k_n^3 \leq 7n$.
1. We start by considering the configuration \( k_1 \leq k_2 \leq \dots \leq k_n \) such that the sum \( \sum k_i \) is maximized under the constraint \( \sum k_i^3 \leq 7n \).

2. First, we analyze the smallest possible value for \( k_1 \). If \( k_1 \geq 2 \), then:
   \[
   \sum k_i^3 \geq n \cdot 2^3 = 8n > 7n
   \]
   This violates the given constraint. Therefore, \( k_1 = 1 \).

3. Next, we consider the largest possible value for \( k_n \). If \( k_n \geq 3 \), we can replace the pair \((k_1, k_n) = (1, k_n)\) with \((2, k_n - 1)\). This replacement keeps the sum \( \sum k_i \) the same but decreases \( \sum k_i^3 \) because:
   \[
   1^3 + k_n^3 = 1 + k_n^3 \quad \text{and} \quad 2^3 + (k_n - 1)^3 = 8 + (k_n - 1)^3
   \]
   Since \( 1 + k_n^3 > 8 + (k_n - 1)^3 \) for \( k_n \geq 3 \), the sum of cubes decreases.

4. By repeating this process, we can achieve a configuration consisting only of 1's and 2's. Let \( p \) be the number of 2's and \( n - p \) be the number of 1's. We need to maximize:
   \[
   \sum k_i = 2p + (n - p) = n + p
   \]
   under the constraint:
   \[
   \sum k_i^3 = 8p + (n - p) \leq 7n
   \]
   Simplifying the constraint:
   \[
   8p + n - p \leq 7n \implies 7p \leq 6n \implies p \leq \frac{6n}{7}
   \]

5. The maximum value of \( p \) is \( \left\lfloor \frac{6n}{7} \right\rfloor \). Therefore, the maximum value of the sum \( \sum k_i \) is:
   \[
   n + \left\lfloor \frac{6n}{7} \right\rfloor
   \]

Conclusion:
\[
\boxed{n + \left\lfloor \frac{6n}{7} \right\rfloor}
\]