In a team based shooter: +1 elimination point awarded for shared kill. Given elimination score of everyone in team, how can we determine its legality? For an elimination, all team members involved in the kill get one point. If there are n players and we are provided with the elimination score of each player, how do we determine if this sequence of n numbers is valid? 
For example, for a team of 5, if the sequence is (6, 1, 1, 0, 2), we can immediately tell this is impossible. Player 1 can't get 6 shared kills since the whole of rest of the team didn't get 6 kills put together. But this one is easy to call. 
Consider a team of 3 with sequence (3, 4, 5). This requires some work. Since we are supposed to tell if it is possible only, let's assume that at most 2 people shared a kill. 
We can build a matrix $\{x_{ij}\}$ where $x_{ij}$ refers to the kills by player $i$ along with $j$. It can be seen that $x_{ii}=0$ and $x_{ij}=x_{ji}$. Since all eliminations are assumed to have only 2 people, kills by 1 are the sum of kills shared with 2 and shared with 3. That is,
$$ x_{12} + x_{13} = 3$$
Similarly, we will have,
$$ x_{21} + x_{23} = 4 $$
$$ x_{31} + x_{32} = 5 $$
By using the fact that the matrix is symmetric, we can solve to get $x_{12} = 1, x_{13} = 2, x_{23} = 3$. So, assuming at most 2 share a kill, we have found a solution. That's ok since we are simply asked a yes or a no. 
But in general, how do we know if the sequence is valid? We will have all sorts of terms such as $x_{1,2345}$ (eliminations by 1 shared with 2,3,4 and 5) etc which we may have to consider to know the answer. 
Any ideas? This problem occurred to me while pondering over the score board of a game in which our team lost horribly. Help will be greatly appreciated. 
 A: For a tl;dr, read the yellow quoted section.
First notice that for a given sequence of scores, you can rearrange them to be in non-decreasing order since the order of the players does not matter to this question. Also notice that once ordered this way we can ignore any terms that are $0$ since they cannot contribute to our system of checking validity. For ex. $(6, 1, 1, 0, 2)$ can be written as $(1, 1, 2, 6)$. This makes the analysis simpler so for the rest of this question assume $ 0 <x_i \leq x_{i+1}$ for any given sequence.
Another thing to notice is that the shared kill system means given a sequence of scores $x_1, x_2, \ldots x_k$, at every step we can subtract $1$ from any two or more players with a non-zero score. A valid sequence is one that can be reduced to all $0$'s for each player by applying this system of subtraction.
Now let us start with small examples and generalize. 
Case: 1 player:
In this case, the only valid score is a $0$ since all kills have to be shared. 
Case: 2 players:
Since at any time we can subtract 1 from two or more players with nonzero scores, it is fairly evident that the only valid set of scores with two players is when their scores are equal i.e. the scores are $(x, x)$
Case: 3 players:
Consider a set of scores $(x_1, x_2, x_3)$ and remember we have rearranged such that $x_1 \leq x_2 \leq x_3$. Now if $x_2 = x_3$, then we can conclude this is valid since we can subtract $x_1$ from both $x_2$ and $x_3$ to give $(0, x', x')$ where $x' = x_2 - x_1 = x_3 - x_1$ and this is just the valid 2 player case since we can ignore the first 0.
If however $x_2 < x_3$ i.e. they aren't equal, then if we can subtract enough of $x_1$ from $x_3$ to get the last two terms equal, we will have reduced to the previous situation in the paragraph above. So the condition we need is that 
$$
x_3 - x_1 \leq x_2 \\
x_3 \leq x_1 + x_2
$$
In other words, with this condition, we can subtract some $k$ from both $x_1$ and $x_3$ such that our sequence is reduced to $(x_1', x_2, x_2)$ where $x_1' = x_1 - k \geq 0$. This is valid since this is just the situation in the first paragraph of the 3 player case.
Doing the examples for case 4 and so on, we get to the following claim for $n$ players where $n \geq 3$

Claim: Given a sequence of non-decreasing scores $x_1, x_2, \ldots, x_n$, the sequence is valid if we have $x_n \leq  \sum_{i=1}^{n-1}x_i$.

We prove the claim by induction. We have shown why this is true for $n=3$ where we needed $x_3 \leq x_1 + x_2$. Now assume for $n$ players, the sequence is valid if $x_n \leq  \sum_{i=1}^{n-1}x_i$. We will prove that for $n+1$ players, the sequence is valid if $x_{n+1} \leq \sum_{i=1}^{n}x_i$.
Consider a non-decreasing sequence of $n+1$ players $x_1, x_2, \ldots x_{n+1}$. If  we have the case that the sequence $x_2, x_3, \ldots x_{n+1}$ is a valid $n$ player sequence i.e. 
$$
x_{n+1} \leq \sum_{i=2}^{n}x_i
$$
then we are done because we can subtract $x_1$ from all of these terms to get the score set $(0, \ x_2 - x_1, \ x_3 - x_1, \  \ldots \ x_{n+1} - x_1)$ where we can ignore the first $0$ and end up with a valid $n$ player sequence.
If we do not have the previous case, but can reduce the sequence to the previous case, we have a valid sequence. Thus if we can subtract enough of $x_1$ from $x_{n+1}$, say $k$ amount where $k < x_1$, such that $x_2, x_3, \ldots x_{n+1}-k$, is a valid $n$ player sequence i.e.
$$
x_{n+1} - k \leq \sum_{i=2}^{n}x_i
$$
then we have reduced ourselves to the previous case. For such a $k$ to exist (since $k \leq x_1$), we need the condition that:
$$
\begin{align}
x_{n+1} - x_1 &\leq \sum_{i=2}^{n}x_i \\
x_{n+1} &\leq \sum_{i=2}^{n}x_i + x_1 \\
x_{n+1} &\leq \sum_{i=1}^{n}x_i  \\
\end{align}
$$
where the last line is what was required for the induction.
Sorry this is a long post but I wanted to be clear. Let me know if I can clarify anything or have messed up somewhere.
