There are too many possible interpretations of the rules:
Case 1: You are not allowed to bid twice in a row
Perfect play: The best Option is not to bid at all, because bidding 5$ will lose you money.
Explanation: It is hard to put a starting point to analyse possible strategies because of infinite payrolls. But if every player plays a perfect strategy and can predict all other players, we have absolutely no chance involved and the outcome of each round can be predicted perfectly beforehand. Therefore there is no risky gamble, but always an expected Win/Loss.
In this scenario nobody will ever bid 21, because when you bid 21 you are guaranteed to loose at least 1\$ so whatever strategy would lead you to bid 21 eventually would never be taken in the first place, because not bidding at all is better than loosing.
This means you can always safely bet 20 without loosing or winning anything, because no one will ever bid 21 in perfect play. What about 19 ? If you bet 19, you can be sure that the guy who did bet 18 will now bet 20, because 20 is safe and will give him 0 loss (and nobody will ever bet 21, because nobody will use a loosing strategy leading him there) So betting 19 is a guaranteed loss.
This in case makes 18 to a perfectly safe bid, since no one will ever bet the sure loss of 19. And by induction we can follow this road downwards. Nobody will ever bet 17, since the next guy will pick 18 and you can't pick 19 because 19 is a sure loss. So all odd numbers are a guaranteed loss. So all perfect players will only ever bet even numbers.
So if the starting bid would be 4 or 6 (or any even number) the first one to bid would win. But so nobody will bid 5\$ since it is a guaranteed loss.
Case 2: One bidder can bid multiple times in a row and has to pay his last 2 bids if he wins
So if I bid 5\$ and then immediately overbid myself with 6\$ and no one else bids, I would get the 20 but would have to pay 5 and 6 (highest and second highest bid) so my net win would be 9\$.
In this case someone can always try to jump an odd number, by bidding twice in a row and get a winning even number. So the first bid would be 5+6 (since paying both bids is only 11 and you get 20 net win 9\$) next save bid is 7+8 and then 9+10. Nobody will bet 11, since 11+12 (or any other two bets in a row over 10) will result in a net loss. So the first one to bet 10 wins. So the question is how the bidding system actually works. But everyone will start bidding up to 10, or nobody will bid at all, depending on the number of your chances to get the two bets in a row. So with two people it is still attractive to try and get the bets 9+10 (if we assume chances 50/50 to get the 9\$ and 50/50 to get the 10\$, expected net results are 0,-9,1,10 so an expected winning of 0.5\$)
Case 3: One bidder can bid multiple times in a row, but whoever else is highest below him has to pay the extra
So if Geoffrey bids 5\$ and I bid 6,7 and 8 afterwards and no one else bids, I would get the 20\$ for 8 and Geoffrey would loose his 5\$ as second highest bidder.
This will also result in a race to the top, cut off depending on the chances of getting you bet in (so the number of people)
Case X: Humans involved with unpredictable bids
Don't bet at all, or try psychological tricks to intimidate them like at the poker-table. But this is too unpredictable, best to not bid at all... Or probably bid 5\$ and try to intimidate everyone away from bidding and if someone bids higher, let them destroy each other.