This is a partial answer (see note 2 below), but mostly 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 $(\mathbb{Z},\max,+)$-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 max 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), or $-\infty$ if there is no accepting path at all. The language $L(A)$ defined by the automaton $A$ is the set of words having nonnegative multiplicity (i.e., such that there exists an accepting path with $v_1+\cdots+v_n \geq 0$).
[Note on edit: I had initially written this for automata with weights in the integer ring $(\mathbb{Z},+,\times)$ (so we take the sum of the $v_1\cdots v_n$). But I realize, after re-reading the question, that the indended meaning is for the weights to be in the tropical semiring $(\mathbb{Z}\cup\{-\infty\},\max,+)$ (we can forget about the weight $-\infty$ by simply omitting the transition). I'm sorry if this may have caused further confusion.]
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 $(\mathbb{Z},\max,+)$-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 $D$?
Question 4 (added following JDH's comment): Is there an algorithm which, given $A$ and $D$, with $D$ deterministic, decides whether $L(A) = L(D)$ holds? [Update: see note 2 below.]
Note that if we consider the question of unweighted automata instead (i.e., all weights are assumed to be $0$), then the answer to 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”). The answer to question 4 is also “yes” in this case (the argument proceeds by first determinizing as just explained, then minimizing the resulting automata, and the simply comparing them). But the fact that we can have negative multiplicities completely changes things.
Note 2: the answer to question 4 (for $(\mathbb{Z},\max,+)$-weighted automata) is negative, in fact even if $D$ is the trivial automaton which accepts every word (one state, both initial and finite, and no transition) so that $L(D) = \sigma^*$, there is no algorithm to decide whether $L(A) = \sigma^*$. See Almagor, Boker & Kupferman, “What's decidable about weighted automata?”, theorem 4.1 (taking care that since they use $(\mathbb{Z}\cup\{+\infty\},\min,+)$ as tropical semiring instead of $(\mathbb{Z}\cup\{-\infty\},\max,+)$ as above, the set of words of multiplicity $<1$ in their sense is the set of words of multiplicity $\geq 0$ in the above sense).
