Take a $n \times n$ chess page.Determine the $n$ such that  we can put the numbers $1,2,3, \ldots ,n$ in the squares of the page such that we know the following two  conditions are true:

a) for each row we know all the numbers $1,2,3, \ldots ,n$ have appeared on it and the numbers that are in the black squares of that row have the same sum as the sum of the numbers in the white squares of that row.

b) for each column we know all the numbers $1,2,3, \ldots ,n$ have appeared on it and the numbers that are in the black squares in that column  have the same sum as the sum of the numbers in the white squares of that column.
To solve this problem, we need to determine the values of \( n \) for which we can place the numbers \( 1, 2, 3, \ldots, n \) on an \( n \times n \) chessboard such that the following conditions hold:

a) For each row, all the numbers \( 1, 2, 3, \ldots, n \) appear exactly once, and the sum of the numbers in the black squares is equal to the sum of the numbers in the white squares.

b) For each column, all the numbers \( 1, 2, 3, \ldots, n \) appear exactly once, and the sum of the numbers in the black squares is equal to the sum of the numbers in the white squares.

1. **Coloring the Chessboard:**
   - Consider a standard \( n \times n \) chessboard where the squares are alternately colored black and white.
   - If \( n \) is even, each row and each column will have an equal number of black and white squares, i.e., \( \frac{n}{2} \) black squares and \( \frac{n}{2} \) white squares.

2. **Sum of Numbers in Each Row and Column:**
   - The sum of the numbers \( 1, 2, 3, \ldots, n \) is given by:
     \[
     S = \frac{n(n+1)}{2}
     \]
   - For the sums of the black and white squares to be equal in each row and column, the sum of the numbers in the black squares must be equal to the sum of the numbers in the white squares. Therefore, each sum must be:
     \[
     \frac{S}{2} = \frac{n(n+1)}{4}
     \]

3. **Condition for \( n \):**
   - For the sum \( \frac{n(n+1)}{4} \) to be an integer, \( n(n+1) \) must be divisible by 4.
   - Since \( n \) and \( n+1 \) are consecutive integers, one of them is even, and the product \( n(n+1) \) is always even. For \( n(n+1) \) to be divisible by 4, one of the numbers must be divisible by 4.
   - Therefore, \( n \) must be of the form \( 4k \) or \( 4k+3 \).

4. **Odd \( n \):**
   - If \( n \) is odd, the number of black and white squares in each row and column will not be equal, making it impossible to satisfy the condition that the sums of the numbers in the black and white squares are equal.

5. **Conclusion:**
   - The only possible values of \( n \) that satisfy the given conditions are multiples of 4, i.e., \( n = 4k \).

The final answer is \( \boxed{ n = 4k } \) where \( k \) is a positive integer.