In a soccer tournament each team plays exactly one game with all others. The winner gets $3$ points, the loser gets $0$ and each team gets $1$ point in case of a draw. It is known that $n$ teams ($n \geq 3$) participated in the tournament and the final classification is given by the arithmetical progression of the points, the last team having only 1 point.
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[*] Prove that this configuration is unattainable when $n=12$
[*] Find all values of $n$ and all configurations when this is possible
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1. **General Case Analysis:**
   Let \( A_1, A_2, \ldots, A_n \) be the participants of the tournament, and let \( a_i \) denote the number of points scored by participant \( A_i \) such that \( a_1 > a_2 > \cdots > a_n \). The total number of points \( S \) in the tournament is given by:
   \[
   S = a_1 + a_2 + \cdots + a_n
   \]
   Since the points form an arithmetic progression with the last team having only 1 point, we can write:
   \[
   a_i = a_1 - (i-1)d
   \]
   where \( d \) is the common difference. Given that \( a_n = 1 \), we have:
   \[
   a_1 - (n-1)d = 1 \implies a_1 = 1 + (n-1)d
   \]
   The sum \( S \) of the arithmetic progression is:
   \[
   S = \frac{n}{2} \left( a_1 + a_n \right) = \frac{n}{2} \left( (1 + (n-1)d) + 1 \right) = \frac{n}{2} \left( 2 + (n-1)d \right) = n + \frac{n(n-1)d}{2}
   \]

2. **Points Distribution:**
   Each game between two teams contributes at least 2 points to the total sum \( S \). The total number of games is \( \binom{n}{2} = \frac{n(n-1)}{2} \). Therefore:
   \[
   S \geq 2 \cdot \frac{n(n-1)}{2} = n(n-1)
   \]
   Substituting the expression for \( S \):
   \[
   n + \frac{n(n-1)d}{2} \geq n(n-1)
   \]
   Simplifying, we get:
   \[
   n + \frac{n(n-1)d}{2} \geq n(n-1) \implies 2n + n(n-1)d \geq 2n(n-1) \implies n(n-1)d \geq 2n(n-2)
   \]
   Dividing by \( n(n-1) \):
   \[
   d \geq 2 \left( \frac{n-2}{n-1} \right)
   \]
   For large \( n \), \( \frac{n-2}{n-1} \approx 1 \), so \( d \geq 2 \).

3. **Case \( d = 1 \):**
   If \( d = 1 \), then:
   \[
   a_i = n - i + 1
   \]
   The sum \( S \) is:
   \[
   S = \sum_{i=1}^{n} (n-i+1) = \frac{n(n+1)}{2}
   \]
   We know:
   \[
   \frac{n(n+1)}{2} \geq n(n-1) \implies n(n+1) \geq 2n(n-1) \implies n+1 \geq 2n-2 \implies 3 \geq n
   \]
   Since \( n \geq 3 \), the only possible value is \( n = 3 \).

4. **Case \( d \geq 3 \):**
   If \( d \geq 3 \), then:
   \[
   a_1 = 1 + (n-1)d \geq 1 + 3(n-1) = 3n - 2
   \]
   The maximum points a team can get is \( 3(n-1) \):
   \[
   a_1 \leq 3(n-1)
   \]
   Therefore:
   \[
   3n - 2 \leq 3(n-1) \implies 3n - 2 \leq 3n - 3 \implies -2 \leq -3
   \]
   This is a contradiction, so \( d \geq 3 \) is not possible.

5. **Case \( d = 2 \):**
   If \( d = 2 \), then:
   \[
   a_i = 2n - 2i + 1
   \]
   The sum \( S \) is:
   \[
   S = \sum_{i=1}^{n} (2n - 2i + 1) = n^2
   \]
   We also have:
   \[
   3a + 2b = n^2 \quad \text{and} \quad a + b = \frac{n(n-1)}{2}
   \]
   Solving these equations, we get:
   \[
   3a + 2 \left( \frac{n(n-1)}{2} - a \right) = n^2 \implies 3a + n(n-1) - 2a = n^2 \implies a = n
   \]
   Therefore, \( b = \frac{n(n-1)}{2} - n \).

6. **Checking \( n = 12 \):**
   For \( n = 12 \):
   \[
   S = 12^2 = 144
   \]
   \[
   a + b = \frac{12 \cdot 11}{2} = 66
   \]
   \[
   3a + 2b = 144
   \]
   Solving these, we get:
   \[
   a = 12 \quad \text{and} \quad b = 54
   \]
   This configuration is not possible because the points do not form an arithmetic progression with the last team having only 1 point.

7. **Possible Values of \( n \):**
   Checking \( n = 3, 4, 5, 6 \):
   - \( n = 3 \): Possible.
   - \( n = 4 \): Possible.
   - \( n = 5 \): Not possible.
   - \( n = 6 \): Not possible.

The final answer is \( \boxed{ n = 3 } \) and \( n = 4 \).