Greedy algorithm :p Consider the sequence $a_1=1$ $a_{n+1}=2a_n+a_{n-1}.......a_1+1$ or $a_{n+1}=b_n+.....b_1+1$ and $b_n=a_1+....a_n+1$ and chose the set $A$ as $+a_i$; $-b_i$

The answer is yes. An example is given by $A = \{(-1)^n \cdot F_n|n\geq 2,n \in \mathbb{N}\} = \{1,-2,3,-5,8,-13,21,\ldots\}$, where $F_n$ are the Fibonnaci-numbers. Clearly $A$ is sumfull, since for every $n \geq 2$, $(-1)^n \cdot F_n = (-1)^{n+1} \cdot F_{n+1} + (-1)^{n+2} \cdot F_{n+2}$, because this comes down to $F_n = -F_{n+1}+F_{n+2}$, which is trivial. Before we continue, let us recall two identities about Fibonacci-numbers: Identity 1: For every $n \geq 1$ it holds $\sum_{i=1}^{n} F_{2i} = F_{2n+1}-1$. Identity 2: For every $n \geq 2$ it holds $\sum_{i=2}^{n} F_{2i-1} = F_{2n}-1$. Both are easily proved by induction. Now we prove that every positive integer can be written as the sum of the elements of a finite subset of $A$. We will prove for every $n \geq 2$ that we can write every integer between 1 and $F_{2n+1}-1$ only using elements $a \in A$ with $|a| < F_{2n+1}$. Firstly note that this holds for $n=2$, since $1 = 1$, $2 = 3+1-2$, $3=3$ and $4=3+1$. Now suppose the statement is true for $n=k$. Consider any integer $m$ with $F_{2k+1} \leq m < F_{2k+3}$. If $m < F_{2k+2}$, we can write $m = F_{2k+2} + \overline{m}$, with $-F_{2k+1} < \overline{m} < 0$. This implies $\overline{m} = -F_{2k+1} + m'$, where $m'$ is a number with $1 \leq m' < F_{2k+1}$, so by induction hypothesis $m'$ can be expressed by elements of $A$ with absolute value smaller then $F_{2n+1}$. Let $B$ be the set of these elements. Now $m = F_{2k+2} - F_{2k+1} + \text{The sum of the elements of} B$. For this $m$ the statement is now true. Furthermore, suppose $m = F_{2k+2}$, then we write $m = F_{2k+2}$ and the statement is true. Finally, if $F_{2k+2} < m < F_{2k+3}$, we write $m = F_{2k+2} + m'$, with $0 < m' < F_{2k+1}$, and again, by induction hypothesis, $m'$ can be expressed by elements of $A$ with absolute value smaller then $F_{2n+1}$ and the statement is true again. So by induction, since the Fibonnaci-numbers are unbounded, every positive integer can be expressed as the sum of the elements of a finite subset of $A$. The proof that every negative number can be written in this way is almost the same. So we only have to prove that $0$ can't be written as the sum of the elements of a subset of $A$. Suppose it can, and let $B \subset A$ be the subset such that the sum of the elements of $B$ is zero. Define $B^- = \{b \in B|b < 0\}$ and $B^+ = \{b \in B|b > 0\}$. Let $b_1 = \max{B^+}$ and $b_2 = \min{B^-}$. W.l.o.g. assume $|b_1| > |b_2|$ (the other case is the same). Define $B^-_1 = \{|b| | b \in B^-\}$. Now we know that the sum of the elements of $B^-_1$ equals the sum of the elements of $B^+$. Furthermore, the sum of the elements of $B^+$ is at least $b_1 = F_{2n}$ for some positive integer $n$. If $n=1$, $B^-_1$ is empty, a contradiction. So $n \geq 2$. By definition of $A$ and $B^-_1$, $B^-_1$ consists of numbers $F_{2i-1}$ with $2 \leq i \leq n$ (since $|b_1| > |b_2|$). Now, the sum of the elements of $B^-_1$ is at most $\sum_{i=2}^n F_{2i-1} = F_{2n}-1$ by Identity 2. This is a contradiction with the fact that the sum of the elements of $B^+$ and $B^-_1$ has to be equal, since one of them is at least $F_{2n}$ and the other one is smaller than or equal to $F_{2n}-1$. The case $|b_2| > |b_1|$ can be handled in the same way. Now we have proven that $A = \{(-1)^n \cdot F_n|n\geq 2,n \in \mathbb{N}\}$ fulfills every condition.

Please help me: In Daniel's solution, how does |$b_1$| > |$b_2$| imply that $B^{-}_1$ consists of the numbers $F_{2i-1}$ with 2 $\leq$ i $\leq$ n? I am not getting how the max-min condition of $B^{+} and B^{-}$ leads to the bound (contradiction against there being a sum of elements equal to 0), though I do understand what's going on until the n $\geq$ 2 part.