Show that an even integer in $\mathbb Z_{2n}$ can be written twice as a sum of disjoint terms of a sequence (EDITED) Question: Let $n \ge 2$ be an even integer, and let $S = (g_1,\dots,g_{\lfloor \log_2(n) \rfloor + 2})$ be a sequence with $|S| = \lfloor \log_2(n) \rfloor + 2$ terms (not necessarily distinct) over $\mathbb Z_{2n}$. Suppose that no term is $0$ or $n \pmod {2n}$.

Show that there exist an even integer $2m \in \mathbb Z_{2n}$ which can be written twice as a sum of disjoint terms of $S$.

EDIT: Show that EITHER there exist an even integer $2m \in \mathbb Z_{2n}$ which can be written twice as a sum of disjoint terms of $S$ OR there exist a subsequence whose sum of the terms is $0 \pmod {2n}$.
My attempt: I got to show that there exist an integer which can be written twice as a sum of disjoint terms of $S$, but I was not able to show that this integer can be taken as even. Indeed, let $u$ be the number of even integers in $S$ and $v$ be the number of odd integers in $S$. One can form $2^u \left[ \binom{v}{0} + \binom{v}{2} + \binom{v}{4} + \dots + \binom{v}{2 \lfloor v/2 \rfloor} \right] = 2^{u+v-1}  = 2^{\lfloor \log_2(n) \rfloor + 1} > n$ even distinct subsums. By Pigeonhole Principle, there exist two of them which are equal modulo $2n$, but they are not necessarily disjoint. If we cancel common terms, it is possible that the even sums become odd sums... That's the problem.
A similar argument shows that there exist two disjoint subsequences of $S$ which are congruent modulo $2n$, but I have no idea how this can be useful since we don't know the parity after cancel.
In addition, I didn't use the fact that $n$ is even.
PS: If $S$ has two identical even integers then we are trivially done (these identical terms can be seen as disjoints, since it appears twice in the sequence).
 A: Step 1: Suppose we need to select a set of $k$ distinct numbers of the set of the first $m$ natural numbers.  What is the most we can choose such that no two subsets have the same sum mod $m$?
It is trivial to show that we can never choose $m$.  We therefore want to choose as many as we can from $1$ to $m-1$
For $m=3$, We can choose both $1$ and $2$
For $m=4$: We can choose no more than $2$ options, for example $1$ and $2$
For $m=5$: We can choose $1, 2$ and $4$
It is trivial to show that we can always safely select $1, 2, 4, ..., 2^{\lfloor\log_2(m)\rfloor-1}$ and have no two subsets share a sum mod m.  This is a total of $\lfloor\log_2(m)\rfloor$ safe selections.
Suppose that we can do better, and can select $\lfloor\log_2(m)\rfloor+1$ options from the numbers $1,2,...,m-1$.  In order for no two of our subsets to have the same sum, the total number of subsets must be less than or equal to the number of possible sums.  The number of subsets from a set of $Z$ objects is $2^Z$, which means that if $Z=\lfloor\log_2(m)\rfloor+1$ then $2^{\lfloor\log_2(m)\rfloor+1}$ must be less than or equal to $m$.  If we define an integer $t$ such that $2^t\leq m<2^{2+1}$, then $2^{\lfloor\log_2(m)\rfloor+1}=2^{t+1}>m$, which means that we can't do such a selection
Step 2: Suppose we need to select $k$ even numbers from the set of the first $w=2m$ natural numbers.  Based on what we did in Step 1, we have a total of $\lfloor\log_2(m)\rfloor=\lfloor\log_2(2m)\rfloor-1=\lfloor\log_2(w)\rfloor-1$ that we can choose
Step 3: Suppose that we can also select odd numbers, and can't have subsets with the same even sum.
If we select $j$ odd numbers, we can choose an upper bound of $\lfloor\log_2(w)\rfloor-1$ even numbers for a total of $\lfloor\log_2(w)\rfloor-1+j$ numbers
However, every pair of odd numbers selected is an even number that it is as if we selected, which means that our maximum number of choices for any given $j\geq2$ is $\lfloor\log_2(w)\rfloor-1+j-\binom{j}{2}$.  Because $\binom{j}{2}\geq j$ for $j\geq3$, we only need to consider the cases of choosing up to 2 odd numbers.
In essence, we can at most select $\lfloor\log_2(w)\rfloor$ numbers in order to not have any even numbers that we can express as the sum in two different ways.  This is because if $j=0$, we get $\lfloor\log_2(w)\rfloor-1$, if $j=1$, we get $\lfloor\log_2(w)\rfloor$, and if $j=2$ we get $\lfloor\log_2(w)\rfloor$.
Step 4: We need to prove that if we choose $\lfloor\log_2(n)\rfloor+2$ numbers from the set $1,2,3,...,n-2,n-1,n+1,n+2,...,2n-1$, we by necessity have two different subsets that share an even sum.
Consider our expression $\lfloor\log_2(w)\rfloor$.  We defined $w$ as an even number, so we can redefine $w=2n$, and $\lfloor\log_2(w)\rfloor=\lfloor\log_2(2n)\rfloor=1+\lfloor\log_2(n)\rfloor$
However, we are choosing 1 more than that, and therefore we must have at least two subsets that have the same even sum
