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.