As a mathematician, I can further confirm that, mathematically, more balanced teams will be formed with an even number of participants.

Let's assume there are n players in a tournament. The organizers decided to form m teams (captains) from n players. According to the tournament rules, the teams must have the same number of people, so each team has k people (baskets), and n = m * k (all numbers are integers). Question: with what number of people in a team (baskets) are the teams most balanced?

Each player has a certain rating Rn. Since the victory of players from any basket is equivalent, we define the strength of each team P[m] through the total rating of the team P[m] = SUM R(i = 1 ... k).

The total number of people n is divided into k baskets, and each i-th basket has a rating spread D(D= R[(i-1)*k] - R[i*k]).

Let's consider a specific case: the teams are formed by 2 out of m captains - the top-1 captain and the last-ranked captain.

The team of the last-ranked captain will have the following strength:
P1 = R[m] (captain's rating) + R[m+1] (since he is the first, he chooses the next-ranked player) + R[3*m] (he chooses last, i.e., after m-1 captains from this basket) + R[3*m+1] (he chooses first) + R[5*m] (chooses last) + ... (depends on the parity of the teams).
If the number of baskets k is odd, then this team has (k-1)/2 players who are top-1 in their baskets and 1+(k-1)/2 players who are losers in their baskets.
If the number of teams is even, then this team has (k-1)/2 players who are top-1 in their baskets and (k+1)/2 players who are losers in their baskets.

The strength of the team of the captain with the highest rating:
P2 = R1 (captain's rating) + R[2*m] (chooses last) + R[2m+1] (chooses first) + R[4*m] (last) + R[4*m+1] (chooses first) + ...
In this team, the situation with the top-1 and losers is reversed compared to the previous one, equal to 1+(k-1)/2 and (k-1)/2, respectively.

Let's find the difference in strength between teams 1 and 2:

P2 - P1 = (R[1] + R[2*m] + R[2*m+1] + R[4*m] + R[4*m+1] + ...) - (R[m] + R[m+1] + R[3*m] + R[3*m+1] + R[5*m] + ...)

Let's reorganize the expression, comparing the ratings of players from the same basket:

P2-P1 = (R[1] - R[m]) + (R[2*m] - R[m+1]) + (R[2*m+1] - R[3*m]) + (R[4*m] - R[3*m+1]) + (R[4*m+1] - R[5*m]) + ...

The differences in parentheses are nothing more than the rating spreads, that is:

P2-P1 = D[1] + (-D[2]) + D[3] + (-D[4]) + D[5] + ...

Thus, indeed, the parity of the teams affects the team's strength. If the number of people in the team is even:

P2-P1 = D[1] + (-D[2]) + D[3] + (-D[4])

then the spread of team strengths is less.

However, let's remember about the rating spread in the baskets. Obviously, the rating spread in the basket will decrease if we reduce the number of people in the basket, while maintaining the total number of participants, in other words, increasing the number of baskets. This can be clearly demonstrated by pushing the situation to the absurd: N people participate, let's make N/2 baskets of 2 people. The difference in ratings between pairs will be minimal, the spread tends to zero:

P2-P1 = D[1] + (-D[2]) + D[3] + (-D[4]) + D[5] + ... -> 0

the strengths of the teams are almost equal.

In this case, two options for balancing the teams are considered - making an even number or increasing the number of baskets. As shown above, each of the options balances the teams in its own way. But the option of an even number of baskets is associated with a big problem of the previous tournament - draws.
As a result, I am "for" increasing the number of baskets to 5.

P.S. if it is really needed, then it is possible to quantitatively assess the differences in team strengths by writing a program that calculates the strength of teams with different numbers of baskets, based on the actual rating of the players who registered for the tournament