In each square of the table below, we must write a different integer from $1$ to $17$, such that the sum of the numbers in each of the eight columns is the same, and the sum of the numbers in the top row is twice the sum of the numbers in the bottom row. Which number from $1$ to $17$ can be omitted? 

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1. **Sum of integers from 1 to 17**:
   The sum of the first \( n \) positive integers is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
   For \( n = 17 \):
   \[
   S = \frac{17 \times 18}{2} = 153
   \]

2. **Sum of the numbers in the top and bottom rows**:
   Let the sum of the numbers in the bottom row be \( x \). According to the problem, the sum of the numbers in the top row is \( 2x \). Therefore, the total sum of the numbers in the top and bottom rows is:
   \[
   2x + x = 3x
   \]

3. **Sum of the numbers in all columns**:
   Since the sum of the numbers in each of the eight columns is the same, the total sum of the numbers in the table must be a multiple of 8. Therefore, \( 3x \) must be a multiple of 8.

4. **Finding \( x \)**:
   We need to find \( x \) such that \( 3x \) is a multiple of 8. The possible values of \( x \) that satisfy this condition are:
   \[
   x = 8k \quad \text{for integer } k
   \]
   Since \( 3x \) must be a multiple of 8, we can write:
   \[
   3x = 8m \quad \text{for integer } m
   \]

5. **Modulo calculation**:
   The total sum of the numbers from 1 to 17 is 153. We need to find a number \( y \) from 1 to 17 that can be omitted such that the remaining sum is a multiple of 8. Let \( S' \) be the sum of the remaining numbers:
   \[
   S' = 153 - y
   \]
   We need \( S' \) to be a multiple of 8:
   \[
   153 - y \equiv 0 \pmod{8}
   \]
   Simplifying, we get:
   \[
   153 \equiv 1 \pmod{8}
   \]
   Therefore:
   \[
   1 - y \equiv 0 \pmod{8}
   \]
   \[
   y \equiv 1 \pmod{8}
   \]
   The number \( y \) that satisfies this condition is:
   \[
   y = 9
   \]

Conclusion:
The number that can be omitted is \( \boxed{9} \).