Counting Possible Sums in a Group of Integers Say you have 5 buckets.
Each bucket contains a distinct number $[1, 10]$.
So why can it be guaranteed that the total sum of 1 or more numbers from the 5 buckets will equal the total sum of 1 or more OTHER numbers from the 5 buckets?
Example:
6, 7, 8, 9, 10.
We know that $6+9=15$ and $7+8= 15$.  Regardless of the distinct numbers we fill the buckets with, we will always be able to find cases where this condition holds.
Another Example:
1,2,3,5,10 We know $1+2= 3$ and $3 = 3$
I find this extremely interesting but I'm not quite sure how to explain this observation.
In my reasoning, I know the lowest value that can be obtained is 1, and the highest, $10+9+8+7=34$ (assuming we must leave one number for the other group).
What principles explain simply why this works?
 A: It is the pigeonhole principle with a bit of work to utilize it.  There are $31$ nonempty sets of buckets.  The sums of subsets can range from $1$ to $40$.  If the range were only $1$ to $30$ we would be done because we would know that two sets had the same sum.  If $1$ is a possible sum, the maximum sum is $35$.  In any case there at most $35$ possible sums.  Then note that if $1$ is included, you can't have neighboring numbers at the top end, so the maximum sum is $10+8+6+4+1=29$ and there must be two sums the same, here $10+4=8+6$.  Now we know that $1$ is not in a bucket.  If $2$ is in a bucket, the maximum sum is $10+9+6+5+2=32$ and there must be two that match, here $10+5=9+6$ or $10+8+7+6+2=33$ and we have $8=6+2$ and otherwise the range is too small.  Now we know neither $1$ nor $2$ is in a bucket, so see what happens if you try $3$.  Soon you will always have two pairs of neighboring numbers and you can use that for your sum.
A: This is the pigeonhole principle at work, although it takes a little digging to get it to work. 
Firstly we can always assume that the numbers are different, or we could just match two that are identical. So although you gave me that the numbers were distinct, that's actually not needed to make this work.
So given the smallest value chosen, $s$, the largest sum possible is $s+7+8+9+10  = s+34$, giving a range of $35$ possible sums, although the all-five-numbers can't possibly match anything.
Now the analysis has to devolve into cases, to get the range small enough that the pigeonhole principle applies, of more possible subset sums than the range that contains them. Discarding the empty set and the five-number set, there are 30 subset sums of interest. 
Looking at the top four numbers, we review cases to drive them below $30$ by attempting to avoid a common sum [noting that $s$ cannot equal a difference in the proposed top-$4$ set]:
Sum: $34$, combination: $7,8,9,10$ - $7+10=8+9$
Sum: $33$, combination: $6,8,9,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+10=9+6$
Sum: $32$, combination (1): $5,8,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $32$, combination (2): $6,7,9,10$ - $6+10=7+9$
Sum: $31$, combination (1): $4,8,9,10$ - $s$ cannot be any of $1,2$  and $s=3$ gives $3+10=4+9$
Sum: $31$, combination (2): $5,7,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $31$, combination (3): $6,7,8,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+8=6+7$ 
Sum: $30$, combination (1): $3,8,9,10$ - $s$ cannot be any of $1,2$
Sum: $30$, combination (2): $4,7,9,10$ - $s$ cannot be any of $1,2,3$
Sum: $30$, combination (3): $5,6,9,10$ - $5+10=6+9$
Sum: $30$, combination (4): $5,7,8,10$ - $5+10=7+8$
Sum: $30$, combination (5): $6,7,8,9$ - $6+9=7+8$  
And otherwise the top numbers sum below $30$ and the pigeonhole principle guarantees matching sums.
