Let $ n \geq 3$ be an odd integer. We denote by $ [\minus{}n,n]$ the set of all integers greater or equal than $ \minus{}n$ and less or equal than $ n$.
Player $ A$ chooses an arbitrary positive integer $ k$, then player $ B$ picks a subset of $ k$ (distinct) elements from $ [\minus{}n,n]$. Let this subset be $ S$.
If all numbers in $ [\minus{}n,n]$ can be written as the sum of exactly $ n$ distinct elements of $ S$, then player $ A$ wins the game. If not, $ B$ wins.
Find the least value of $ k$ such that player $ A$ can always win the game.
1. **Define the problem and the set:**
   Let \( n \geq 3 \) be an odd integer. We denote by \([ -n, n ]\) the set of all integers greater than or equal to \(-n\) and less than or equal to \(n\). This set contains \(2n + 1\) elements.

2. **Player A and Player B's actions:**
   - Player A chooses an arbitrary positive integer \( k \).
   - Player B picks a subset \( S \) of \( k \) distinct elements from \([ -n, n ]\).

3. **Winning condition for Player A:**
   Player A wins if all numbers in \([ -n, n ]\) can be written as the sum of exactly \( n \) distinct elements of \( S \).

4. **Determine the least value of \( k \):**
   To ensure that Player A can always win, we need to find the minimum \( k \) such that any subset \( S \) of \( k \) elements can be used to express every number in \([ -n, n ]\) as the sum of exactly \( n \) distinct elements from \( S \).

5. **Consider the number of sums:**
   - The number of possible sums of \( n \) distinct elements from a set of \( k \) elements is given by the binomial coefficient \( \binom{k}{n} \).
   - We need this number to be at least \( 2n + 1 \) to cover all elements in \([ -n, n ]\).

6. **Calculate the minimum \( k \):**
   - We need \( \binom{k}{n} \geq 2n + 1 \).
   - For large \( k \) and \( n \), \( \binom{k}{n} \approx \frac{k^n}{n!} \).
   - To ensure \( \frac{k^n}{n!} \geq 2n + 1 \), we solve for \( k \).

7. **Approximate \( k \):**
   - For simplicity, assume \( k \approx n + 1 + \frac{n-1}{2} \).
   - This approximation comes from considering the smallest \( k \) such that the sum of the smallest \( k \) numbers can cover the range \([ -n, n ]\).

8. **Verify the solution:**
   - If \( k = n + 1 + \frac{n-1}{2} \), then \( k = n + 1 + \frac{n-1}{2} = n + 1 + \frac{n}{2} - \frac{1}{2} = \frac{3n}{2} + \frac{1}{2} \).
   - Since \( k \) must be an integer, we take the ceiling of this value: \( k = \left\lceil \frac{3n}{2} + \frac{1}{2} \right\rceil \).

The final answer is \( \boxed{ k = \left\lceil \frac{3n}{2} + \frac{1}{2} \right\rceil } \).