Pass on your radiation
Python 2, score:  56 50  48 bytes - 6 = 42
This is more an attempt to allow to remove more than 1 or 2 characters than to achieve the lowest possible score. Given the scoring scheme, allowing one more byte to be removed this way would cost at least one extra byte, leaving the score unchanged at best.
You may remove the 3rd to the 8th character, i.e. any digit of \$545454\$.
n=545454;x=input();print x[:n%11]+x[n%11+n%9/4:]

Try it online!
How?
Removing the \$n\$th digit from \$545454\$ and applying a modulo \$11\$ results in \$n+1\$. Applying a modulo \$9\$ instead allows to figure out whether the number was changed or not.
 pattern |    x   | x mod 9 | x mod 11
---------+--------+---------+----------
  545454 | 545454 |    0    |   (8)
  _45454 |  45454 |    4    |    2
  5_5454 |  55454 |    5    |    3
  54_454 |  54454 |    4    |    4
  545_54 |  54554 |    5    |    5
  5454_4 |  54544 |    4    |    6
  54545_ |  54545 |    5    |    7