Existence of a set Does there exist $T \subset \mathbb{Q}$ such that, for every $z \in \mathbb{Z}$, there exists a unique finite set $A_{z} \subset T$ such that $z=\sum_{t \in A_{z}}{t}$?
 A: Let's agree that $0\notin \Bbb N$.
Let $z_n$, $n\in\Bbb N$, be an enumeration of $\Bbb Z$.
For $n\in \Bbb N$, let $a_n=3^{-n}$, $A=\{\,a_n\mid n\in\Bbb N\,\}$,  $b_n=z_n-\sum_{k=1}^na_k$, $B=\{\,b_n\mid n\in \Bbb N\,\}$.
Note that $0<\sum a_k<\frac12$ so that the map $n\mapsto b_n$ is injective (and of course $n\mapsto a_n$ is also injective) and $A\cap B=\emptyset$.
Now let 
$$T=A\cup B.$$ 
Clearly, we can write every integer $\in\Bbb Z$ as sum of elements of $T$, namely $$\tag1z_n=b_n+\sum_{k=1}^na_n.$$
On the other hand, let $U\subset T$ be a finite subset with $s:=\sum_{u\in U}u\in \Bbb Z$.
As $U\subset A$ would imply $0<s<1$, we conclude that $U\cap B$ is non-empty.
Let $n$ be maximal with $b_n\in U$.
We have $3^{n-1}u\equiv 0\pmod 1$ for all $u\in U\cap B$, except that $3^{n-1}b_n\equiv -\frac13\pmod 1$. We conclude that $U\not\subset B$.
Let $m$ be maximal with $a_m\in U$.
If $m<n$, then $3^{n-1}u\equiv 0\pmod 1$ for all $u\in U\cap A$, hence $3^{n-1}\sum_{u\in U} u\equiv-\frac13\pmod 1$, contradiction.
Likewise, if $m>n$, then $3^{m-1}\sum_{u\in U} u\equiv \frac13\pmod 1$, contradiction.
We conclude that $m=n$.
Now we show by induction on $n$ ($=m$) that we have a representation as in $(1)$, i.e., that $s\in\Bbb Z$ implies $$U=\{b_n\}\cup\{\,a_k\mid 1\le k\le n,\}.$$ The case $n=1$ follows immediately from $n=m$.
So let $n>1$ and the claim is known to hold for smaller $n$.
Assume first that $b_{n-1}\in U$. Then $3^{n-2}u\equiv 0\pmod 1$ for all $u\in U$, except that $3^{n-2}a_n\equiv \frac19$, $3^{n-2}b_{n-1}\equiv -\frac13$, $3^{n-2}b_{n}\equiv -\frac49$, and - possibly - $3^{n-2}a_{n-1}\equiv \frac13$. That makes $\sum u-\frac23$ or $\equiv-\frac13$, contradiction.
Thus $b_{n-1}\notin U$.
Then the set  $U':=U\setminus\{a_n,b_n\}\cup \{b_{n-1}\}$ produces the sum $s'=s-a_n-b_n+b_{n-1}=s-z_n+z_{n-1}\in\Bbb Z$. By induction hypothesis, $U'=\{b_{n-1}\}\cup\{\,a_k\mid 1\le k\le n-1\,\}$, and the claim for $U$ follows. $\square$

Remark. The above represents $0$ as a non-empty sum, namely as a sum of $n+1$ summands where $z_n=0$. If we allow the empty sum, we thus obtain two representations of $0$. Under that convention, just let $z_n$ be an enumeration of $\Bbb Z\setminus\{0\}$
