A = \(\left(1-\dfrac{4}{a^2}\right)\left(1-\dfrac{4}{b^2}\right)\)

    = \(\dfrac{\left(a-2\right)\left(a+2\right)\left(b-2\right)\left(b+2\right)}{a^2b^2}\) 

    = \(\dfrac{ab\left(a+2\right)\left(b+2\right)}{a^2b^2}\)

   = \(\dfrac{ab+2\left(a+b\right)+4}{ab}\)

  = \(1+\dfrac{8}{ab}\)

 \(\le1+\dfrac{8}{\left(\dfrac{a+b}{2}\right)^2}=1+8=9\)

Let  be the sum of all the  numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most  tables. So  can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then  increases. Since  has finitely many possible values,  can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative.