Find the smallest natural number nonzero n so that it exists in real numbers $x_1, x_2,..., x_n$ which simultaneously check  the conditions:
1) $x_i \in [1/2 , 2]$ , $i = 1, 2,... , n$ 
2)  $x_1+x_2+...+x_n \ge \frac{7n}{6}$
3) $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}\ge \frac{4n}{3}$
To find the smallest natural number \( n \) such that there exist real numbers \( x_1, x_2, \ldots, x_n \) satisfying the given conditions, we will analyze the inequalities step by step.

1. **Given Conditions:**
   - \( x_i \in \left[\frac{1}{2}, 2\right] \) for \( i = 1, 2, \ldots, n \)
   - \( x_1 + x_2 + \cdots + x_n \ge \frac{7n}{6} \)
   - \( \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} \ge \frac{4n}{3} \)

2. **Analyzing the Reciprocal Inequality:**
   Since \( x_i \in \left[\frac{1}{2}, 2\right] \), we can derive an upper bound for \( \frac{1}{x_i} \):
   \[
   \frac{1}{x_i} \le \frac{5}{2} - x_i
   \]
   This is derived from the quadratic inequality:
   \[
   0 \le (2 - x_i)\left(x_i - \frac{1}{2}\right) = \frac{5x_i}{2} - 1 - x_i^2
   \]
   Rearranging, we get:
   \[
   \frac{1}{x_i} \le \frac{5}{2} - x_i
   \]

3. **Summing the Inequalities:**
   Summing over all \( n \) terms, we get:
   \[
   \frac{4n}{3} \le \sum_{i=1}^n \frac{1}{x_i} \le \sum_{i=1}^n \left(\frac{5}{2} - x_i\right)
   \]
   Simplifying the right-hand side:
   \[
   \sum_{i=1}^n \left(\frac{5}{2} - x_i\right) = \frac{5n}{2} - \sum_{i=1}^n x_i
   \]
   Using the second condition \( \sum_{i=1}^n x_i \ge \frac{7n}{6} \), we get:
   \[
   \frac{4n}{3} \le \frac{5n}{2} - \frac{7n}{6}
   \]
   Simplifying the right-hand side:
   \[
   \frac{5n}{2} - \frac{7n}{6} = \frac{15n}{6} - \frac{7n}{6} = \frac{8n}{6} = \frac{4n}{3}
   \]

4. **Equality Conditions:**
   For the inequalities to hold as equalities, we must have:
   \[
   \frac{1}{x_i} = \frac{5}{2} - x_i
   \]
   Solving for \( x_i \):
   \[
   \frac{1}{x_i} = \frac{5}{2} - x_i \implies x_i^2 - \frac{5x_i}{2} + 1 = 0
   \]
   Solving this quadratic equation:
   \[
   x_i = \frac{5 \pm 1}{4} = \frac{3}{2} \text{ or } 1
   \]

5. **Finding the Minimum \( n \):**
   To satisfy both conditions, we need to find the smallest \( n \) such that:
   \[
   \sum_{i=1}^n x_i = \frac{7n}{6} \quad \text{and} \quad \sum_{i=1}^n \frac{1}{x_i} = \frac{4n}{3}
   \]
   Let \( x_i = \frac{1}{2} \) for \( p \) terms and \( x_i = 2 \) for \( n - p \) terms. Then:
   \[
   \frac{p}{2} + 2(n - p) = \frac{7n}{6}
   \]
   Simplifying:
   \[
   \frac{p}{2} + 2n - 2p = \frac{7n}{6} \implies 2n - \frac{3p}{2} = \frac{7n}{6}
   \]
   Multiplying through by 6:
   \[
   12n - 9p = 7n \implies 5n = 9p \implies n = \frac{9p}{5}
   \]
   The smallest integer \( n \) is when \( p = 5 \):
   \[
   n = \frac{9 \times 5}{5} = 9
   \]

The final answer is \( \boxed{9} \).