Find all finite sets $M \subset R, |M|  \ge  2$, satisfying the following condition: 
[i]for all $a, b \in  M, a \ne b$, the number $a^3 - \frac{4}{9}b$ also belongs to $M$.
[/i]
(I. Voronovich)
1. **Establishing the size of the set \( M \):**
   - Let \( M = \{x_1, x_2, \ldots, x_n\} \) with \( x_1 < x_2 < \cdots < x_n \) and \( n \geq 3 \).
   - We claim that \( x_1^3 - \frac{4}{9} x_n = x_1 \). Assume the contrary, that \( x_1^3 - \frac{4}{9} x_n > x_1 \).
   - By the property of \( M \), we have:
     \[
     \left\{ x_1^3 - \frac{4}{9} x_n, x_1^3 - \frac{4}{9} x_{n-1}, \ldots, x_1^3 - \frac{4}{9} x_2 \right\} \subset M.
     \]
   - Since:
     \[
     x_1 < x_1^3 - \frac{4}{9} x_n < x_1^3 - \frac{4}{9} x_{n-1} < \cdots < x_1^3 - \frac{4}{9} x_2 \leq x_n,
     \]
     it follows that:
     \[
     x_1^3 - \frac{4}{9} x_n = x_2, \quad x_1^3 - \frac{4}{9} x_{n-1} = x_3, \ldots, x_1^3 - \frac{4}{9} x_2 = x_n.
     \]
   - In particular, we obtain:
     \[
     x_3 - x_2 = \frac{4}{9} (x_n - x_{n-1}),
     \]
     and:
     \[
     x_n - x_{n-1} = \frac{4}{9} (x_3 - x_2).
     \]
   - These two equations are contradictory unless \( x_n = x_{n-1} \), which by assumption is not the case. Hence, \( x_1^3 - \frac{4}{9} x_n = x_1 \).

2. **Similar argument for \( x_n \):**
   - A similar argument reveals that \( x_n^3 - \frac{4}{9} x_1 = x_n \).

3. **Considering intermediate elements:**
   - Consider the numbers:
     \[
     x_2^3 - \frac{4}{9} x_n < x_3^3 - \frac{4}{9} x_n < \cdots < x_{n-1}^3 - \frac{4}{9} x_n.
     \]
   - These numbers are all lower bounded by \( x_1^3 - \frac{4}{9} x_n \) and upper bounded by \( x_n^3 - \frac{4}{9} x_1 \). Thus, they must coincide with \( x_2, \ldots, x_{n-1} \). In particular, we conclude \( x_2^3 - \frac{4}{9} x_n = x_2 \).

4. **Further contradiction:**
   - Consider the numbers:
     \[
     x_2^3 - \frac{4}{9} x_{n-1} < x_3^3 - \frac{4}{9} x_{n-1} < \cdots < x_{n-2}^3 - \frac{4}{9} x_{n-1} < x_n^3 - \frac{4}{9} x_{n-1}.
     \]
   - These numbers are all bounded below by \( x_1^3 - \frac{4}{9} x_n = x_1 \) and bounded above by \( x_n^3 - \frac{4}{9} x_1 = x_n \). Hence, they coincide with \( x_2, \ldots, x_{n-1} \). In particular, \( x_2^3 - \frac{4}{9} x_{n-1} = x_2 \).
   - But we have found above that \( x_2^3 - \frac{4}{9} x_n = x_2 \) as well. These two statements yield a contradiction. Thus, \( |M| \leq 2 \).

5. **Considering \( |M| = 2 \):**
   - Let \( M = \{a, b\} \) with \( a < b \). By construction, \( a^3 - \frac{4}{9} b < b^3 - \frac{4}{9} a \), and both belong to the set. We thus obtain:
     \[
     a^3 - \frac{4}{9} b = a \quad \text{and} \quad b^3 - \frac{4}{9} a = b.
     \]
   - Solving these equations, we get:
     \[
     a^2 + ab + b^2 = \frac{5}{9},
     \]
     and:
     \[
     a^3 + b^3 = \frac{13}{9} (a + b).
     \]

6. **Case \( a + b = 0 \):**
   - If \( a + b = 0 \), then we obtain the set:
     \[
     \left\{ -\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3} \right\}.
     \]

7. **Case \( a + b \neq 0 \):**
   - If \( a + b \neq 0 \), then:
     \[
     a^2 - ab + b^2 = \frac{13}{9}.
     \]
   - This, together with \( a^2 + ab + b^2 = \frac{5}{9} \), yields:
     \[
     ab = -\frac{4}{9},
     \]
     and thus:
     \[
     a + b = \pm 1 \quad \text{and} \quad a - b = \pm \frac{\sqrt{17}}{3}.
     \]
   - From here, we deduce the following two more sets:
     \[
     \left\{ \frac{1 - \sqrt{17}}{6}, \frac{1 + \sqrt{17}}{6} \right\}, \quad \text{and} \quad \left\{ \frac{-1 - \sqrt{17}}{6}, \frac{-1 + \sqrt{17}}{6} \right\}.
     \]

The final answer is \( \boxed{ \left\{ -\frac{\sqrt{5}}{3}, \frac{\sqrt{5}}{3} \right\}, \left\{ \frac{1 - \sqrt{17}}{6}, \frac{1 + \sqrt{17}}{6} \right\}, \left\{ \frac{-1 - \sqrt{17}}{6}, \frac{-1 + \sqrt{17}}{6} \right\} } \).