Let $n$ be a positive integer.  The numbers $\{1, 2, ..., n^2\}$ are placed in an $n \times n$ grid, each exactly once.  The grid is said to be [i]Muirhead-able[/i] if the sum of the entries in each column is the same, but for every $1 \le i,k \le n-1$, the sum of the first $k$ entries in column $i$ is at least the sum of the first $k$ entries in column $i+1$.  For which $n$ can one construct a Muirhead-able array such that the entries in each column are decreasing?

[i]Proposed by Evan Chen[/i]
To determine for which \( n \) it is possible to construct a Muirhead-able array such that the entries in each column are decreasing, we need to analyze the construction method and verify the conditions.

1. **Case \( n \) is even:**
   - Fill the grid left to right, top to bottom in decreasing order.
   - Reverse all even-numbered rows.
   - This ensures that each column has the same sum and the entries in each column are decreasing.

2. **Case \( n \) is odd and \( n > 3 \):**
   - Let \( n = 2k + 3 \).
   - Take out the middle \( 3n \) numbers to fill the middle 3 rows later.
   - Fill each column from left to right, avoiding the middle 3 cells in each column.
   - For each column, take the top \( k \) numbers and the bottom \( k \) numbers and place them in decreasing order.

3. **Filling the middle 3 rows:**
   - Denote the middle \( 3n \) numbers by \( a+1, a+2, \ldots, a+3n \).
   - Fill the middle 3 rows as follows:
     - First row: \( a+3n, a+3n-1, \ldots, a+2n+1 \).
     - Second row: \( a+n+1, a+n+3, \ldots, a+2n, a+n+2, a+n+4, \ldots, a+2n-1 \).
     - Third row: \( a+(n+1)/2, a+(n-1)/2, \ldots, a+1, a+n, a+n-1, \ldots, a+(n+3)/2 \).

4. **Verification of Muirhead condition:**
   - For \( k \leq k+1 \), the rows are in decreasing order up to this point.
   - For \( k > k+3 \), the cumulative sum of each column from the bottom to top is in increasing order.
   - For \( k = k+2 \), the maximum positive difference between an element and its left neighbor in this row is 2, ensuring the cumulative sum condition.
   - For \( k = k+3 \), the cumulative sum in the reverse direction is non-decreasing due to the maximum positive difference of 1.

5. **Examples:**
   - For \( n = 5 \):
     \[
     \begin{array}{ccccc}
     25 & 24 & 23 & 22 & 21 \\
     20 & 19 & 18 & 17 & 16 \\
     11 & 13 & 15 & 12 & 14 \\
     8 & 7 & 6 & 10 & 9 \\
     1 & 2 & 3 & 4 & 5 \\
     \end{array}
     \]
   - For \( n = 7 \):
     \[
     \begin{array}{ccccccc}
     49 & 47 & 45 & 43 & 41 & 39 & 37 \\
     48 & 46 & 44 & 42 & 40 & 38 & 36 \\
     35 & 34 & 33 & 32 & 31 & 30 & 29 \\
     22 & 24 & 26 & 28 & 23 & 25 & 27 \\
     18 & 17 & 16 & 15 & 21 & 20 & 19 \\
     2 & 4 & 6 & 8 & 10 & 12 & 14 \\
     1 & 3 & 5 & 7 & 9 & 11 & 13 \\
     \end{array}
     \]

The final answer is \( \boxed{ n \neq 3 } \).