Initially, there are $n$ red boxes numbered with the numbers $1,2,\dots ,n$ and $n$ white boxes numbered with the numbers $1,2,\dots ,n$ on the table. At every move, we choose $2$ different colored boxes and put a ball on each of them. After some moves, every pair of the same numbered boxes has the property of either the number of balls from the red one is $6$ more than the number of balls from the white one or the number of balls from the white one is $16$ more than the number of balls from the red one. With that given information find all possible values of $n$
1. **Understanding the Problem:**
   - We have \( n \) red boxes numbered \( 1, 2, \ldots, n \) and \( n \) white boxes numbered \( 1, 2, \ldots, n \).
   - Each move involves placing a ball in one red box and one white box.
   - After some moves, for each pair of boxes with the same number, the number of balls in the red box is either 6 more than the number of balls in the white box, or the number of balls in the white box is 16 more than the number of balls in the red box.

2. **Setting Up the Equations:**
   - Let \( r_i \) be the number of balls in the red box \( i \) and \( w_i \) be the number of balls in the white box \( i \).
   - The conditions given are:
     \[
     r_i = w_i + 6 \quad \text{or} \quad w_i = r_i + 16
     \]
   - Rearranging the second condition, we get:
     \[
     r_i = w_i - 16
     \]

3. **Summing the Balls:**
   - Summing over all \( i \) from 1 to \( n \), we have:
     \[
     \sum_{i=1}^n r_i = \sum_{i=1}^n (w_i + 6) \quad \text{or} \quad \sum_{i=1}^n r_i = \sum_{i=1}^n (w_i - 16)
     \]
   - Let \( k \) be the number of pairs where \( r_i = w_i + 6 \) and \( n - k \) be the number of pairs where \( r_i = w_i - 16 \).

4. **Calculating the Total Balls:**
   - For the first condition:
     \[
     \sum_{i=1}^k r_i = \sum_{i=1}^k (w_i + 6) = \sum_{i=1}^k w_i + 6k
     \]
   - For the second condition:
     \[
     \sum_{i=k+1}^n r_i = \sum_{i=k+1}^n (w_i - 16) = \sum_{i=k+1}^n w_i - 16(n - k)
     \]
   - Combining these, we get:
     \[
     \sum_{i=1}^n r_i = \sum_{i=1}^n w_i + 6k - 16(n - k)
     \]

5. **Equating the Total Balls:**
   - Since the total number of balls in red and white boxes must be equal (each move adds one ball to a red and one to a white box):
     \[
     \sum_{i=1}^n r_i = \sum_{i=1}^n w_i
     \]
   - Therefore:
     \[
     \sum_{i=1}^n w_i + 6k - 16(n - k) = \sum_{i=1}^n w_i
     \]
   - Simplifying, we get:
     \[
     6k - 16(n - k) = 0
     \]
     \[
     6k - 16n + 16k = 0
     \]
     \[
     22k = 16n
     \]
     \[
     \frac{k}{n} = \frac{16}{22} = \frac{8}{11}
     \]

6. **Finding \( n \):**
   - For \( k \) to be an integer, \( n \) must be a multiple of 11.
   - Therefore, \( n = 11m \) for some \( m \in \mathbb{N} \).

The final answer is \( \boxed{ n = 11m } \) for some \( m \in \mathbb{N} \).