Sets of integers in which every element is the sum of two others

i) Does there exists a nonempty subset of integers $S$, such that for all $a \in S$, there exists $b,c \in S$ such that $a = b + c$, where $a, b, c$ are distinct integers.

Edited to add: ii) Can an $S$ with these properties be finite, and also have the property that $a \in S$ implies $-a \notin S$

This question is inspired by another question, which has not been fully answered yet.

Yes, you can take the set of integers between $-n$ and $n$ inclusive for any $n \ge 3$. Here $0 = 1 + (-1)$, $1 = 2 + (-1)$ and $x = (x - 1) + 1$ for $x \ge 2$. Similar formulae for negative values.

EDIT: Answer for new version of the question Set found by hand:

$$ \{ -22, -20, -18, -16, -14, -12, -10, -2, 1, 3, 7, 8, 15, 23 \} $$

EDIT: Answer with 8 elements. If you add -12 to this set you will get set with odd number of elements. $$ \{ -10, -8, -6, -2, 1, 3, 4, 5 \} $$

I posted an answer to this question previously, giving a finite set. That answer was incorrect: this is a different answer, which contains lower bounds on finite sets of the sort you ask for.

A set S such that (∀a ∈ S)(∃b,c ∈ S):(b≠c  and  a = b + c) we will call self-supporting: a self-supporting set S which is disjoint from −S will be strongly self-supporting.

Proposition 1. A self-supporting set must contain at least three positive elements, and three negative elements.

[Proof] A self-supporting set S is non-empty; because the property of self-support is preserved by negation of the elements, without loss of generality we may consider sets containing positive elements.

The largest positive element of S can only be formed as a sum of two other positive elements; thus it has at least three positive elements 0<c<b<a such that a = b + c. The smallest positive element cannot be formed as a sum of smaller positive elements, so S contains negative elements. By similar arguments, it must contain at least three negative elements.

As the set {−3, −2, −1, 1, 2, 3} is self-supporting, this bound is optimal.

Proposition 2. A strongly self-supporting set must contain at least four positive elements, and four negative elements.

[Proof] Suppose that S is self-supporting, but only has three positive elements 0<c<b<a. Then a = b + c. We cannot form b as a sum of c with some other positive element of S; it then follows that b can only be formed as a sum of a and −c, which implies that S is not strongly self-supporting. Thus, a strongly self-supporting set contains at least four positive elements; and similarly for negative elements.

The smaller set in flagar's answer, {−10, −8, −6, −2, 1, 3, 4, 5}, is thus a minimal-size strongly self-supporting set --- and may also be the set having elements of the smallest absolute value.