Details of the vote counting procedure:

We will use the voting method known as “instant runoff”, modified because many voters are also candidates. No voter’s ballot will be counted while she is still in the running as a candidate.  

Selection begins by eliminating one pair at a time in turn, starting with the pair with the fewest first place votes.  If two or more pairs are tied for fewest first place votes then the tie will be broken by number of second place votes and so on. Once a pair is eliminated, their ballots will be considered in the ensuing iterations. The eliminated pair will also be removed from all the ballots so pairs behind them move up a place on each ballot.  This process will be repeated from the beginning until there are only two pairs left, with the pair with the higher number of first place votes designated for selection.

After the first pair is selected, the process will be repeated for the second pair, and so on (with the now selected pair’s ballots in consideration and that pair’s name removed from all ballots).

In the event a tie cannot be broken by then comparing the non-excluded votes for second place, then third place, etc., we will go to a second tie-breaking procedure.  This secondary tie-breaker will use the more traditional  ranking system among non-excluded ballots to determine the selected pair (akin to college sports polls).   

Here is a fuller explanation, followed by an actual example of the process detailed:

Again, we make the determination of only one invitee at a time.   If we need more than one name, the entire process is repeated as necessary.

Say we have four pairs remaining in consideration (to get down to four pairs, it would be the same method as in this illustration).   Ballots which any of these four pairs have submitted are not counted at this point.   Similarly, any names other than these four are excluded from the ballots we are counting --- each ballot would, in effect, now rank these four pairs from top to bottom in one of the four spots (if one or more teams were not ranked on the full ballot, then they are not listed (see round 4 below).  

If we have 24 remaining ballots after excluding the ballots of these four pairs, we then determine which of these pairs has the fewest ballots with the top ranking among only these four pairs.  If there is a tie for the fewest first place votes, then we eliminate the pair with the fewest second place votes. This pair is now eliminated (for this single invite determination), and if either of the pair had submitted a ballot, their ballot(s) would now be added back into the mix, while their name is removed from the 24-26 now considered ballots, with each ballot now effectively naming up to three pairs.

Of those three remaining pairs, we eliminate the pair with the lowest number of top votes among those three, and again add their ballots back into the mix, while removing their names from the ballot for the consideration of the remaining pairs.   Now we are left with two pairs, and the pair with more votes placing them above the other pair, gets the invite.  

If we need to determine one or more additional invitees, we repeat the process from an entirely fresh start, though with the now invited pair's name eliminated from all ballots, and their ballots re-added for consideration of the other additional invitees.

Example of the process, assuming we are down to seven pairs with initially eight non-excluded ballots (i.e. ballots cast by those not still remaining in consideration):

Round 1: Eliminate G, who has no 1st place votes.

Voter 1   ABCDEFG
Voter 2   BCDEFGA
Voter 3   BDFACEG
Voter 4   CDBAGFE
Voter 5   CEGABDF
Voter 6   DAGFECB
Voter 7   EABCDFG
Voter 8   FAGEDCB

Round 2: Eliminate A, D, E, or F, who have only one 1st place vote each. F is eliminated, with no second place votes.   Since G was been eliminated in round 1, any ballot(s) cast by G’s pair are now considered (there may be zero, one, or two ballots cast by G’s pair --- for illustrative purposes only, we are always assuming eliminated pairs cast exactly one ballot).

Voter 1   ABCDEF
Voter 2   BCDEFA
Voter 3   BDFACE
Voter 4   CDBAFE
Voter 5   CEABDF
Voter 6   DAFECB
Voter 7   EABCDF
Voter 8   FAEDCB

Round 3: Eliminate D or E, who have only one 1st place vote each and two second place votes each.  E is eliminated, with only 1 third place vote.

Voter 1   ABCDE
Voter 2   BCDEA
Voter 3   BDACE
Voter 4   CDBAE
Voter 5   CEABD
Voter 6   DAECB
Voter 7   EABCD
Voter 8   AEDCB
Voter G  CABDE
Voter F   BACED

Round 4: Eliminate D, who has only one 1st place vote.  Voter E did not list D on their ballot in the top ten spots.

Voter 1   ABCD
Voter 2   BCDA
Voter 3   BDAC
Voter 4   CDBA
Voter 5   CABD
Voter 6   DACB
Voter 7   ABCD
Voter 8   ADCB
Voter G  CABD
Voter F   BACD
Voter E   ACB

Round 5: Eliminate B or C.   Since B & C each have the same number of first, second, and third place votes, we go to the secondary tie-breaker of total poll points of the original full ballots of voters 1-8, and G, F, E, D.   We will assume B has 96 points and C has 92 points, and we would eliminate C.

Voter 1   ABC
Voter 2   BCA
Voter 3   BAC
Voter 4   CBA
Voter 5   CAB
Voter 6   ACB
Voter 7   ABC
Voter 8   ACB
Voter G  CAB
Voter F   BAC
Voter E   ACB
Voter D   ABC

Round 6:  Pair A has received 8 votes, B 5 votes.  A is selected.

Voter 1   AB
Voter 2   BA
Voter 3   BA
Voter 4   BA
Voter 5   AB
Voter 6   AB
Voter 7   AB
Voter 8   AB
Voter G  AB
Voter F   BA
Voter E   A
Voter D   AB
Voter C   BA

The entire process is now repeated to find any required additional pairs.   A’s ballot is now considered along with voters 1-8, with A removed from those ballots.