Regex will work for you:

import re

def format(n):
    return re.sub(r"(\d)(?=(\d{4})+(?!\d))", r"\1,", str(n))

>>> format(12345)
'1,2345'
>>> format(12345678)
'1234,5678'
>>> format(123456789)
'1,2345,6789'

Explanation:

Match:

• (\d) Match a digit...
• (?=(\d{4})+(?!\d)) ...that is followed by one or more groups of exactly 4 digits.

Replace:

• \1, Replace the matched digit with itself and a ,

You can break your number into chunks of 10000's using modulus and integer division, then str.join using ',' delimiters

def commas(n):
    s = []
    while n > 0:
        n, chunk = divmod(s, n)
        s.append(str(chunk))
    return ','.join(reversed(s))

>>> commas(123456789)
'1,2345,6789'
>>> commas(123)
'123'

Sounds like a locale thing^(*). This prints 12,3456,7890 (Try it online!):

import locale

n = 1234567890

locale._override_localeconv["thousands_sep"] = ","
locale._override_localeconv["grouping"] = [4, 0]
print(locale.format_string('%d', n, grouping=True))

That's an I guess hackish way based on this answer. The other answer there talks about using babel, maybe that's a clean way to achieve it.

^(*) Quick googling found this talking about Chinese grouping four digits, and OP's name seems somewhat Chinese, so...