A $ 3 \times n$ grid is filled as follows. The first row consists of the numbers from $ 1$ to $ n$ arranged in ascending order. The second row is a cyclic shift of the top row: $ i,i\plus{}1,...,n,1,2,...,i\minus{}1$ for some $ i$. The third row has the numbers $ 1$ to $ n$ in some order so that in each of the $ n$ columns, the sum of the three numbers is the same. For which values of $ n$ is it possible to fill the grid in this way? For all such $ n$, determine the number of different ways of filling the grid.
1. **Define the grid and notation:**
   - Let the first row be \( a_i = i \) for \( i = 1, 2, \ldots, n \).
   - Let the second row be a cyclic shift of the first row, denoted as \( b_i \).
   - Let the third row be a permutation of \( 1, 2, \ldots, n \), denoted as \( c_i \).

2. **Sum condition:**
   - The sum of the numbers in each column must be the same. Let this common sum be \( S \).
   - Therefore, for each column \( i \), we have:
     \[
     a_i + b_i + c_i = S
     \]

3. **Calculate the total sum:**
   - The sum of all elements in the first row is:
     \[
     \sum_{i=1}^n a_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}
     \]
   - The sum of all elements in the second row is also:
     \[
     \sum_{i=1}^n b_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}
     \]
   - The sum of all elements in the third row is:
     \[
     \sum_{i=1}^n c_i = \sum_{i=1}^n i = \frac{n(n+1)}{2}
     \]

4. **Total sum of all elements:**
   - The total sum of all elements in the grid is:
     \[
     \sum_{i=1}^n (a_i + b_i + c_i) = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2}
     \]

5. **Common sum \( S \):**
   - Since there are \( n \) columns, the common sum \( S \) must satisfy:
     \[
     nS = \frac{3n(n+1)}{2}
     \]
   - Solving for \( S \):
     \[
     S = \frac{3(n+1)}{2}
     \]

6. **Condition for \( S \) to be an integer:**
   - For \( S \) to be an integer, \( \frac{3(n+1)}{2} \) must be an integer.
   - This implies that \( n+1 \) must be even, i.e., \( n \) must be odd.

7. **Constructing the grid:**
   - Let \( n \) be odd.
   - The first row is \( 1, 2, \ldots, n \).
   - The second row is a cyclic shift of the first row.
   - The third row must be a permutation such that the sum of each column is \( \frac{3(n+1)}{2} \).

8. **Number of ways to fill the grid:**
   - The number of ways to choose the cyclic shift for the second row is \( n \).
   - The number of ways to choose the permutation for the third row is \( n! \).
   - Therefore, the total number of ways to fill the grid is:
     \[
     n \cdot n!
     \]

The final answer is \( \boxed{ n \cdot n! } \) for odd \( n \).