the answer is m = 1011.

claim1: $m \leq 1011$ here is a construction (in attachments) let 2021 vertices be 2021 cities, 2 vertices are connected if they have a direct flights, color each company's flight by a different color, we can see that we can choose at most 1011 cities such that they don't have all company's direct flights claim2: for $m = 1011$, we always choose an incomplete group $d_1,d_2,...,d_{2021}$ respectively number direct flights of company 1,...,2023 we have $2021.d_i =< d_1 +... + d_{2021} =< \frac{2021.2020}{2}$ so $d_i \leq 1010$ let $d_i$ be the company i has the smallest number of flights the number of cities have direct flights of company i is $\leq 2d_i$ so there are at least $2021 - 2d_i$ cities don't have direct flights of company i We always choose $d_i$ cities, which have direct flights of company i, such that they don't have a direct flight of company i between them. So we can choose an incomplete group of at least $2021 - 2d_i + d_i = 2021 - d_i \geq 1011$ cities.