Answer in one line:

''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))

or even shorter starting with Python 3.6 using random.choices():

''.join(random.choices(string.ascii_uppercase + string.digits, k=N))

A cryptographically more secure version: see this post

''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))

In details, with a clean function for further reuse:

>>> import string
>>> import random
>>> def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
...    return ''.join(random.choice(chars) for _ in range(size))
...
>>> id_generator()
'G5G74W'
>>> id_generator(3, "6793YUIO")
'Y3U'

How does it work ?

We import string, a module that contains sequences of common ASCII characters, and random, a module that deals with random generation.

string.ascii_uppercase + string.digits just concatenates the list of characters representing uppercase ASCII chars and digits:

>>> string.ascii_uppercase
'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
>>> string.digits
'0123456789'
>>> string.ascii_uppercase + string.digits
'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'

Then we use a list comprehension to create a list of 'n' elements:

>>> range(4) # range create a list of 'n' numbers
[0, 1, 2, 3]
>>> ['elem' for _ in range(4)] # we use range to create 4 times 'elem'
['elem', 'elem', 'elem', 'elem']

In the example above, we use [ to create the list, but we don't in the id_generator function so Python doesn't create the list in memory, but generates the elements on the fly, one by one (more about this here).

Instead of asking to create 'n' times the string elem, we will ask Python to create 'n' times a random character, picked from a sequence of characters:

>>> random.choice("abcde")
'a'
>>> random.choice("abcde")
'd'
>>> random.choice("abcde")
'b'

Therefore random.choice(chars) for _ in range(size) really is creating a sequence of size characters. Characters that are randomly picked from chars:

>>> [random.choice('abcde') for _ in range(3)]
['a', 'b', 'b']
>>> [random.choice('abcde') for _ in range(3)]
['e', 'b', 'e']
>>> [random.choice('abcde') for _ in range(3)]
['d', 'a', 'c']

Then we just join them with an empty string so the sequence becomes a string:

>>> ''.join(['a', 'b', 'b'])
'abb'
>>> [random.choice('abcde') for _ in range(3)]
['d', 'c', 'b']
>>> ''.join(random.choice('abcde') for _ in range(3))
'dac'

You're almost there. I've just tweaked it a little - you could adjust it as is:

import random

numcount = 5
fstring = ""
length_of_code_you_want = 12
position_to_keep_constant_1 = 2 # first position to keep constant is position 3, index 2
position_to_keep_constant_2 = 6 # 2nd position to keep constant is position 7, index 6
constant_1 = "J" # the value of the constant at position 3
constant_2 = "L" # the value of the constant at position 7

for num in range(length_of_code_you_want): #strings are 19 characters long
    if random.randint(0, 1) == 1:
        x = random.randint(1, 8)
        x += 96
        fstring += (chr(x).upper())
    elif not numcount == 0:
        x = random.randint(0, 9)
        fstring += str(x)
        numcount -= 1

list_out = list(fstring)
list_out[position_to_keep_constant_1] = str(constant_1)
list_out[position_to_keep_constant_2] = str(constant_2)
string_out = "".join(list_out)
print(string_out)

the Random module uses a cryptographically insecure PRG (Mersenne Twister). You want the Secrets module (documentation here), which uses urandom on Linux, or CryptGenRandom() on Windows. In fact, I believe the following code should work (but have not verified it):

from secrets import choice
import string

def get_random_string(length):
    result_str = ''.join(choice(string.ascii_letters) for i in range(length))
    print(result_str)

If you are implementing things where misimplementation may cost you money in any way, I would suggest learning about practical cryptography before proceeding (as it is very easy to make small mistakes that can be a big pain). There are a variety of books people like, for example Cryptography Engineering.