This is not an answer, but an attempt to rephrase the question into something both meaningful and understandable, so that hopefully someone can answer it.
Let $\sigma$ be a finite set, also known as the alphabet, and $\sigma^*$ the set of words on $\sigma$, i.e., the free monoid on $\sigma$ (consisting of finite sequences of elements of $\sigma$).
A $\mathbb{Z}$-weighted finite automaton on $\sigma$ is given by a finite set $Q$ of states, an element $q_0 \in Q$ known as the initial state, a subset $F \subseteq Q$ known as the accepting states, and a finite subset $\delta \subseteq Q \times \sigma \times \mathbb{Z} \times Q$ known as transitions of the automaton (read $(q,x,v,q') \in \delta$ as “the automaton can jump from state $q$ to state $q'$ while consuming symbol $x$ with multiplicity $v$).
For such an automaton, we define the multiplicity of the word $w = x_1\cdots x_n \in \sigma^*$ to be the sum of the $v_1\cdots v_n$ ranging over all accepting paths, that is, all $(q_1,\ldots,q_n)$ and $(v_1,\ldots,v_n)$ such that $q_n\in F$ and $(q_{i-1},x_i,v_i,q_i)\in\delta$ for $1\leq i\leq n$ (note that $q_0$ is the initial state defined with the automaton). The language $L(A)$ defined by the automaton $A$ is the set of words with nonnegative multiplicity.
Important note: this assumes that if a word has no accepting path whatsoever it is part of the language $L(A)$. This strikes me as an odd definition (but it would be even more bizarre to demand both that there exists an accepting path and that the sum of all multiplicities is $\geq 0$). So OP should clarify whether it is really what was intended.
We say that the automaton is deterministic when $\delta$ is actually a function $Q\times\sigma \to \mathbb{Z}\times Q$ (that is, for all $(q,x) \in Q\times \sigma$ there is a unique $(v,q') \in \mathbb{Z}\times Q$ such that $(q,x,v,q') \in \delta$).
Question 1: Is it true that, for every $\mathbb{Z}$-weighted finite automaton $A$ there is a deterministic one $D$ such that $L(D) = L(A)$?
Question 2: If the answer to question 1 is “no”, is there an algorithm which, given $A$, decides whether there is such a $D$?
Question 3: Is there an algorithm which, given $A$ for which there is such a $D$, returns such a $D$?
Note that if we consider the question of unweighted automata instead (i.e., all weights are assumed to be $1$) and modify the definition of the language to be the set of words whose multiplicity is $>0$, then the answer ot questions 1 and 3 is “yes”: in this setup, to construct $D$ we consider the powerset of the set of states of $A$, and create a transition $(\mathbf{q},x,\mathbf{q}')$ in $D$ when $\mathbf{q}'$ is the set of $q'$ such that the transition $(q,x,q')$ exists in $A$ for some $q\in \mathbf{q}$; the accepting states of $D$ are those that contain some accepting state of $A$, and the initial state of $D$ is the singleton $\{q_0\}$ of the accepting state of $A$. This is a classical construction from automata theory (“determinization”). But the fact that we can have negative multiplicities completely changes things.