Take $k\in \mathbb{Z}_{\ge 1}$ and the sets $A_1,A_2,\dots, A_k$ consisting of $x_1,x_2,\dots ,x_k$ positive integers, respectively. For any two sets $A$ and $B$, define $A+B=\{a+b~|~a\in A,~b\in B\}$.

Find the least and greatest number of elements the set $A_1+A_2+\dots +A_k$ may have.

[i] (Andrei Bâra)[/i]
### Lower Bound

1. **Claim**: The minimum size of the set \( A_1 + A_2 + \dots + A_k \) is \( 1 - k + \sum_{i=1}^k x_i \).

2. **Base Case**: Consider two sets \( A \) and \( B \). The minimum size of \( A + B \) is \( |A| + |B| - 1 \). This is because the smallest element in \( A \) paired with all elements in \( B \) gives \( |B| \) distinct sums. Pairing the remaining \( |A| - 1 \) elements with the largest element in \( B \) ensures that these sums are distinct and greater than any sum involving the smallest element in \( A \).

3. **Inductive Step**: Assume the minimum size of the sum of the first \( m \) sets is \( 1 - m + \sum_{i=1}^m x_i \). For the set \( A_1 + A_2 + \dots + A_{m+1} \), the minimum size is:
   \[
   (1 - m + \sum_{i=1}^m x_i) + x_{m+1} - 1 = 1 - (m + 1) + \sum_{i=1}^{m+1} x_i
   \]
   This completes the induction.

4. **Conclusion**: By induction, the minimum size of \( A_1 + A_2 + \dots + A_k \) is \( 1 - k + \sum_{i=1}^k x_i \).

### Upper Bound

1. **Claim**: The maximum size of the set \( A_1 + A_2 + \dots + A_k \) is \( \prod_{i=1}^k x_i \).

2. **Base Case**: For two sets \( A \) and \( B \), the maximum size of \( A + B \) is \( |A| \cdot |B| \). This is because if \( B \) is an arithmetic sequence with a common difference greater than the difference between the largest and smallest elements in \( A \), then all sums \( a + b \) (where \( a \in A \) and \( b \in B \)) are distinct.

3. **Inductive Step**: Assume the maximum size of the sum of the first \( m \) sets is \( \prod_{i=1}^m x_i \). For the set \( A_1 + A_2 + \dots + A_{m+1} \), the maximum size is:
   \[
   \left( \prod_{i=1}^m x_i \right) \cdot x_{m+1} = \prod_{i=1}^{m+1} x_i
   \]
   This completes the induction.

4. **Conclusion**: By induction, the maximum size of \( A_1 + A_2 + \dots + A_k \) is \( \prod_{i=1}^k x_i \).

The final answer is \( \boxed{ 1 - k + \sum_{i=1}^k x_i } \) for the minimum size and \( \prod_{i=1}^k x_i \) for the maximum size.