Reducing two variables at a time Initially the variables $x_1,x_2,\dots,x_n$ have some nonnegative (real) values, and none of the variables has more than half the total value. We are allowed to choose two variables and decrease both of them by some (real) amount, as long as they remain nonnegative. Can we always keep doing this and end up with all variables being zero?
When some variable is more than half the total value, we cannot do it since the variable will still be positive even if we exhaust all other variables. On the contrary, if one variable is exactly half the total value, it is possible by each time choosing it and any other variable.
 A: First note that one can extend the operation to a "generalized move" involving any number of piles each to be reduced by some amount $d$ (as long as each pile has at least size $d$). For example if the piles are $a,b,c,d,e$ one can successively reduce each pair $(a,b),(b,c),(c,d),(d,e),(e,a)$ by $d/2$ and the result is that all 5 piles have been reduced by $d$ at once.
Next if we place the values in nondecreasing order $x_1,x_2,\cdots x_n$ and it happens that $x_{n-1}=x_n$ we can finish as follows, using the term "block" for a sequence of equal values. Starting from the rightmost block, use the above generalized move to reduce its common value to that of the block to its left. At this point there is a new rightmost block whose length has increased to that of the sum of lengths of the original last two blocks. Now repeat. The lengths of the rightmost blocks are increasing, so one can always go on one more step. At the end one gets to a single block which can then be zeroed out by one generalized move.
Now what does one do if the last block has length $1$? In that case the above "work from right to left by blocks" idea cannot be started, as we are not allowed to reduce just one pile. What we do for this case will be explained in terms of five piles $a,b,c,d,e$ in order of nondecreasing value, but the idea will apply in general. We are now assuming $d<e$ since otherwise the last block has size $2$ or more already treated. For this step we only use pair reduction, not the generalized move. The idea is to use piles $a,b,c$ with pile $e$ (keeping $d$ unchanged) with the goal of reducing $e$ until it becomes $d.$ The amount by which $e$ has to be reduced for this is $e-d.$ So among piles $a,b,c$ we require $a+b+c \ge e-d.$ But that is equivalent to $a+b+c+d+e\ge 2e,$ that is, to the size of pile $e$ being at most half the sum. 
Now we successively use $a,b,c$ as needed, each paired with $e$, decreasing each pair in turn. If it happens that $a\ge e-d$ we only need part of $a.$ If $a<e-d \le a+b$ we first delete all of $a$ with some of $e,$ then turn to using the pair $(b,e).$ I think it's clear we can in this way delete by any amount $x$ between $0$ and $a+b+c,$ so we can successfully achieve $d=e$ as desired and then use the previous "last block at least length two" idea.
A: If your variables sum to an odd number, you will never be able to reduce them all to zero. A simple example:
Let $x_0=5, x_1=7, x_2=9$. Choose the smallest and largest variables and reduce both by 5. Now $x_0=0, x_1 = 7$, and $x_2=4$. Now we can reduce both variables by 4, but $x_1$ will be left at 3.
