Your code has several problems:

First, the return in the wrong place. It is inside the for loop but it should be outside. Next, in the following fragment:

for i in range(len(str)):
    n=str[i]
    newStr=str.replace(n, "*")

the n that you passed as the second argument to your function is being overwritten at every loop step. So if your initial string is "abcabcabcd" and you pass n=3 (a number) as a second argument what your loop does is:

n="a"
n="b"
n="c"
...

so the value 3 is never used. In addition, in your loop only the last replacement done in your string is saved:

n="a"
newStr="abcabcabcd".replace("a", "*") --> newStr = "*bc*bc*bcd"
n="b"
newStr="abcabcabcd".replace("b", "*") --> newStr = "a*ca*ca*cd"
...
n="d"
newStr="abcabcabcd".replace("d", "*") --> newStr = "abcabcabc*"

If you test your function (after fixing the return position) with some strings it seems to work fine:

In [7]: replaceN("abcabcabc", 3)
Out[7]: 'ab*ab*ab*'

but if you do the choice more carefully:

In [10]: replaceN("abcabcabcd", 3)
Out[10]: 'abcabcabc*'

then it is obvious that the code fails and it is equivalent to replace only the last character of your string:

my_string.replace(my_string[-1], "*")

The code given by Eric is working fine:

In [16]: ''.join("*" if i % 3 == 0 else char for i, char in enumerate("abcabcabcd"))
Out[16]: '*bc*bc*bc*'

It replaces positions 3rd, 6th, 9th and so on. It may need some adjustment if you don't want the position 0 being replaced too.

Following works and does not use regular expressions

special_chars = {'.','/','|','?','!','_','"',',','-','@','\n','\\'}

def format_word(w, fill):
    if w[-1] in special_chars:
        return fill*(len(w) - 1) + w[-1]
    else:
        return fill*len(w)


def obscure(string, every=3, fill='#'):
    return ' '.join(
        (format_word(w, fill) if (i+1) % every == 0 else w)
        for (i, w) in enumerate(string.split())
    )

Here are some example usage

In [15]: obscure(string)
Out[15]: 'My dear ##########, do you ########## the nature ## the given ##########?'

In [16]: obscure(string, 4)
Out[16]: 'My dear adventurer, ## you understand the ###### of the given ##########?'

In [17]: obscure(string, 3, '?')
Out[17]: 'My dear ??????????, do you ?????????? the nature ?? the given ???????????'

Check this out:

>>> word = 'Python For All'
>>> new_word = ''.join(character for index, character in enumerate(word) if index%3 != 0)
>>> new_word
'ytonFo Al'

I want to replace every second instance of | with a new line

• Menu "Search" > "Replace" (or Ctrl + H)

• Set "Find what" to (.*?\|.*?)[\|]

• Set "Replace with" to \1\r\n

• Enable "Regular expression"

• Click "Replace All"

  

Before:

Name1|Value1|Name2|Value2|Name3|Value3

After:

Name1|Value1
Name2|Value2
Name3|Value3

Notes:

• The above assumes you are editing a text file with Windows EOLs, \r\n.

• If you are using files with different EOLs you can convert them to Windows EOLs using Menu "Edit" > "EOL Conversion".

• If you aren't working with Windows EOL, and you don't wish to convert them, use the following instead:

  • Use \n instead of \r\n for Unix/OS X EOLs

  • Use \r instead of \r\n for Mac OS (up to version 9) EOLs

Further reading

• Notepad++: A guide to using regular expressions and extended search mode

If you need to replace the 19^th and 20^th character of each line with x if and only if they are , and space respectively, you can do:

sed 's/^\(.\{18\}\), /\1x/'

With most sed implementations (all those compliant to POSIX 2024), that can be made more legible by switching from basic to extented regexp with -E:

sed -E 's/^(.{18}), /\1x/'