Let $n$ be a natural number. The finite sequence $\alpha$ of positive integer terms, there are $n$ different numbers ($\alpha$ can have repeated terms). Moreover, if from one from its terms any we subtract 1, we obtain a sequence which has, between its terms, at least $n$ different positive numbers. What's the minimum value of the sum of all the terms of $\alpha$?
1. **Restate the Problem:**
   We are given a finite sequence $\alpha$ of positive integers with $n$ different integers. If we subtract 1 from any term in $\alpha$, the resulting sequence must still contain at least $n$ different positive integers. We need to find the minimum sum of the terms in $\alpha$.

2. **Initial Observations:**
   - If an integer appears more than twice in the sequence, we can reduce the sum by removing extra occurrences.
   - If an integer $m$ appears twice but $m-1$ does not appear, we can reduce the sum by removing one occurrence of $m$.

3. **Optimal Sequence Construction:**
   - Each integer should appear at most twice.
   - The sequence should be constructed with blocks of consecutive terms appearing twice, connected by terms with a difference of 2.

4. **Example Sequences:**
   - For $n=6$, possible sequences are:
     - $1, 2, 2, 4, 5, 5, 7, 9$ with sum $35$.
     - $1, 3, 5, 7, 9, 11$ with sum $36$.

5. **Block Definitions:**
   - Define blocks $A$ and $B$:
     - $A$ block: sequence of consecutive terms appearing twice.
     - $B$ block: sequence of terms with a difference of 2.

6. **Sum Calculation:**
   - Let $A(x, y)$ be the sum of an $A$ block starting at $x$ and ending at $y-1$.
   - Let $B(x, y)$ be the sum of a $B$ block starting at $x$ and ending at $y-1$.
   - The total sum $S$ can be expressed as:
     \[
     S = A(1, n_1) + B(n_1, n_2) + A(n_2, n_3) + \ldots
     \]

7. **Block Sum Formulas:**
   - For an $A$ block starting at $x$ and ending at $y-1$:
     \[
     A(x, y) = \sum_{i=x}^{y-1} 2i = 2 \left( \frac{(y-1)y}{2} - \frac{(x-1)x}{2} \right) = (y-1)y - (x-1)x
     \]
   - For a $B$ block starting at $x$ and ending at $y-1$:
     \[
     B(x, y) = \sum_{i=x}^{y-1} i = \frac{(y-1)y}{2} - \frac{(x-1)x}{2}
     \]

8. **Minimizing the Sum:**
   - The sequence should alternate between $A$ and $B$ blocks to minimize the sum.
   - For $n = 2k$, the minimum sum is:
     \[
     f(n) = 3k^2 + 2k - 1
     \]
   - For $n = 2k+1$, the minimum sum is:
     \[
     f(n) = 3k^2 + 5k + 1
     \]

The final answer is $\boxed{3k^2 + 2k - 1}$ for $n = 2k$ and $\boxed{3k^2 + 5k + 1}$ for $n = 2k+1$.