Pairwise sums in an ordered list The scenario:
In a game with n players, each player has a in individual score and players are ranked accordingly (P1 is the player/score in 1st place, and Pn is last place). Ties are allowed. 
Next, each pairwise sum of all n scores is computed and the sums are then ordered (S1, S2,... Sm). Note that m = nC2 total positions.
Observations:
Clearly S1=P1+P2 and S2=P1+P3. But, S3 can vary depending on the individual scores.  Similarly, Sm=Pn+Pn-1 and Sm-1=Pn-1+Pn-2. But, Sm-3 can vary.
It could be that there are no other restrictions other than these. But, intuitively, it seems like there must be. So, the question is...
The question:
For an individual player/score, Pi (ranked in kth place as an individual), what is the range of possible positions they could be ranked in as part of a pair? 
It may be better to ask the question in reverse. That is, given any Si=Pj+Pk, what are the possiblities for j and k (as a fucntion of i)? 
I hope that's all clear. Thanks very much for any solutions, insights, or resources you can provide.
 A: I'm not sure the problem is entirely well defined, because of the possibility of ties.  So I will first answer the question on the assumption that there are no ties, and will then give some comments on the case with ties.
We write $(i,j)>(k,l)$ to mean that the total score of players $i$ and $j$ is greater than the total score of players $k$ and $l$.  Then we clearly have
$$(1,2)>(1,3)>\cdots>(1,k)\ ,$$
so the highest pair that player $k$ can occupy is certainly no higher than pair number $k-1$.  And indeed this position is possible, because if player $1$ is way ahead of all the others then the top $n-1$ pairs will be
$$(1,2)\,,\ (1,3)\,,\ldots,\ (1,n)\ .$$
Specifically, this will be true if and only if $(1,n)>(2,3)$, and an example where this occurs is if the scores are
$$2n-3\,,\ n-1\,,\ n-2\,,\ n-3\,,\ldots,\ 1\ .\tag{$*$}$$
We can summarise this argument as follows: the best pair that player $k$ can belong to is that where he is paired with player $1$, and in this case there are still $k-2$ players who will do even better when paired with player $1$.  In the same way, the worst that player $k$ can do is when he is paired with player $n$, and then there are still $n-k-1$ players who will be worse when paired with player $n$.  So the lowest pair that player $k$ can be in is the one $n-k$ places from the bottom; this is achieved by a "dual" table to $(*)$ in which player $n$ is far worse than all the others, for example,
$$2n-3\,,\ 2n-4\,,\ldots,\ n\,,\ n-1\,,\ 1\ .$$
To sum up: the highest pair that player $k$ can occupy is the $(k-1)$th from the top, the lowest is the $(n-k)$th from the bottom.  I will leave it to others to determine whether all intermediate places are actually possible - my guess would be "yes".  For example, if there are $k\ge6$ players with scores
$$2k-2\,,\ k\,,\ k-1\,,\ k-3\,,\ k-4\,,\ k-5\,,\ldots,\ 3\,,\ 2\,,\ 0$$
then the top pairs are
$$(1,2)>(1,3)>\cdots>(1,k-1)>(2,3)>(1,k)>(2,4)\ ,$$
and so player $k$'s best pair is the $k$th from the top.
Note however that even though I have assumed no tied scores between players, there may well be ties between pairs as soon as we depart from the extreme cases considered above.  So to tell if a given position around the "middle" of the list is possible or not, you would need to define some tie-break mechanism.
For the game with ties possible between players, you would need some method of breaking ties even at the top and bottom of the table.  An extreme example would be where every player has the same score and tied pairs are ordered at random.  In this case, any player could occupy a pair in any position.
