Pass on your radiation The task here is pretty simple.  You should write a program that takes a string as input and outputs it (that is a cat program).

Additionally when the \$n\$th byte of your program is removed (without replacement) the augmented program should take a string as input and remove the \$n\$th character.  If the input to your augmented program is less than \$n\$ characters long it should return it as is. For example the python program



x = input()
if "y" == "y":
  print(x)
else:
  print(x[:16] + x[17:])


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will remove the \$17\$th character if its \$17\$th byte is removed (the first y).

Scoring

Your answer's score will be the number of bytes for which removal does not cause the program to remove the corresponding byte.  With better score being lower.

In our example above the total length is 68 bytes, with one byte (the \$16\$th one) augmenting the program correctly.  That leaves a score of 67 bytes.

If the program were modfied to work in two places like so:

x=input()
u="az"
if u!="az":x=x[:13+(u&lt;"c")]+x[14+(u&lt;"c"):]
print(x)


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Where removing the \$14\$th and \$15\$th characters (a and z) work properly.  The score is 66 bytes or 68 - 2, since the program is 68 bytes long but two of those bytes work properly.  

Rules and clarifications


There must be at least one byte which can be removed that causes the program to remove the corresponding character in the input.  This means a cat program on its own is not a valid submission.
Only one byte will be removed at a time.  You do not have to worry about what happens when multiple bytes are removed from your program at the same time.
You should support strings using printable ascii characters in addition to all characters that appear in your program itself.  Input containing anything else is undefined behavior.
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:]

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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