Let $a$ and $b$ be two real numbers and let$M(a,b)=\max\left \{ 3a^{2}+2b; 3b^{2}+2a\right \}$. Find the values of $a$ and $b$ for which $M(a,b)$ is minimal.
1. We start with the function \( M(a,b) = \max\{3a^2 + 2b, 3b^2 + 2a\} \). We aim to find the values of \(a\) and \(b\) that minimize \(M(a,b)\).

2. To find a lower bound for \(M(a,b)\), we use the fact that the maximum of two numbers is at least as large as their average:
   \[
   M(a,b) \geq \frac{3a^2 + 2b + 3b^2 + 2a}{2}
   \]

3. Simplify the expression inside the average:
   \[
   \frac{3a^2 + 2b + 3b^2 + 2a}{2} = \frac{3a^2 + 3b^2 + 2a + 2b}{2}
   \]

4. To make the expression more manageable, we complete the square for both \(a\) and \(b\):
   \[
   3a^2 + 2a = 3\left(a^2 + \frac{2}{3}a\right) = 3\left(a^2 + \frac{2}{3}a + \frac{1}{9} - \frac{1}{9}\right) = 3\left(\left(a + \frac{1}{3}\right)^2 - \frac{1}{9}\right)
   \]
   \[
   3b^2 + 2b = 3\left(b^2 + \frac{2}{3}b\right) = 3\left(b^2 + \frac{2}{3}b + \frac{1}{9} - \frac{1}{9}\right) = 3\left(\left(b + \frac{1}{3}\right)^2 - \frac{1}{9}\right)
   \]

5. Substitute these completed squares back into the average:
   \[
   \frac{3\left(\left(a + \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 3\left(\left(b + \frac{1}{3}\right)^2 - \frac{1}{9}\right) + 2a + 2b}{2}
   \]

6. Combine the terms:
   \[
   \frac{3\left(a + \frac{1}{3}\right)^2 + 3\left(b + \frac{1}{3}\right)^2 - \frac{2}{3} + 2a + 2b}{2}
   \]

7. Notice that \(2a + 2b\) can be rewritten as \(2(a + b)\), and we can further simplify:
   \[
   \frac{3\left(a + \frac{1}{3}\right)^2 + 3\left(b + \frac{1}{3}\right)^2 - \frac{2}{3}}{2}
   \]

8. Separate the terms:
   \[
   \frac{3\left(a + \frac{1}{3}\right)^2 + 3\left(b + \frac{1}{3}\right)^2}{2} - \frac{1}{3}
   \]

9. Since squares of real numbers are always non-negative, the minimum value of \(3\left(a + \frac{1}{3}\right)^2 + 3\left(b + \frac{1}{3}\right)^2\) is 0, which occurs when \(a + \frac{1}{3} = 0\) and \(b + \frac{1}{3} = 0\). Therefore, \(a = -\frac{1}{3}\) and \(b = -\frac{1}{3}\).

10. Substituting \(a = -\frac{1}{3}\) and \(b = -\frac{1}{3}\) into the expression, we get:
    \[
    M\left(-\frac{1}{3}, -\frac{1}{3}\right) = \max\left\{3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right), 3\left(-\frac{1}{3}\right)^2 + 2\left(-\frac{1}{3}\right)\right\}
    \]
    \[
    = \max\left\{\frac{1}{3} - \frac{2}{3}, \frac{1}{3} - \frac{2}{3}\right\} = \max\left\{-\frac{1}{3}, -\frac{1}{3}\right\} = -\frac{1}{3}
    \]

11. Therefore, the minimum value of \(M(a,b)\) is \(-\frac{1}{3}\), and it occurs when \(a = -\frac{1}{3}\) and \(b = -\frac{1}{3}\).

The final answer is \(\boxed{a = -\frac{1}{3}, b = -\frac{1}{3}}\).