Here is one neat way of organising the matches, which is optimal.
First I'll describe a general method that works for any number 2n that is
a power of 2, say 2n=2^k.
Number the players in binary from 0 to 2n-1. These are all k-bit binary numbers. Then arrange k matches, where the teams in the k-th match are determined by the k-th bit in the players' numbers.
This works because:
Any two players have different numbers, and so their numbers will differ in at least one bit position. In those matches they will be on opposite teams. Note also that in every match both teams contain exactly n players as required.
By leaving out some of the players in the matches you can use same team arrangement with any smaller number of players.
To reduce n, leave out any pair of players with complementary numbers (i.e. numbers that differ in every bit). These players are on opposite teams in every match, so removing them keeps the teams of equal size in all matches.
So for $2n$ players this method uses k matches where
Here is the n=11 case explicitly.
22<2^5 so we use 5-bit numbers resulting in 5 matches. The table below shows this. The players a-v are numbered from 5 to 26 (shown vertically in binary) because I have removed 5 pairs from the 32 players that this match schedule could accomodate (0-4 and their complements 27-31).
Now let's prove this is optimal, using the idea given by loopy_walt in the comments below, which is basically the reverse of the steps that were used to construct the teams:
Suppose we have a solution using k matches. Assign each player a k-bit binary number, where the k-th bit is 0 or 1 depending on whether the player is in the first or the second team in the k-th match.
Since this is a solution, every pair of players will play on opposing teams at least once. Therefore, every player will have a number assigned to them that is different from every other number. So they have 2n unique k-bit binary numbers, which is only possible when 2n<=2^k, or equivalently k>=lg(2n). The least possible k is therefore k=ceiling(lg(2n)), which the previously described method achieves.