This is a common error, if you write an independent python file and execute it.

Defining a custom command

Although you can make sure the Django framework is properly loaded, usually the idea is that you define a custom command [doc].

So in your file structure, if app is the application where your Concert is located, you can construct directories and files (here listed in boldface):

someapp/
    management/
        __init__.py
        commands/
            __init__.py
            load_concerts.py
    __init__.py
    models.py
    views.py

Where the __init__.py files are empty. Then your load_concerts.py can look like:

from django.core.management.base import BaseCommand, CommandError
import json
from app.models import Concert

class Command(BaseCommand):
    help = 'Load JSON concert data'

    def add_arguments(self, parser):
        parser.add_argument('concert_file', type=str)

    def handle(self, *args, **options):
        with open(options['concert_file']) as f:
            data = json.load(f)
        for concert in data:
            Concert.objects.create(**concert)

You can the execute the command with:

$ manage.py load_concerts output.json --settings=my_settings

So just like you run the server (well you can see this as a builtin command), you can thus add specific commands).

By defining the filename as a parameter, we can easily load any .json file we want.

An independent Python program

You can define a loose Python file. In that case you need to load the settings, with:

from django.conf import settings
from projectroot import settings as projectsettings
settings.configure(projectsettings)

import django
django.setup()

import json
from app.models import Concert

with open('output.json') as f:
    data = json.load(f)
for concert in data:
    Concert.objects.create(**concert)

Then you can set the PYTHONPATH to the project root, and thus load the specific settings, but I think this will probably result in more pain that to define a command you can easily reuse.