Probably not. It would be enough to prove it for rationals, since then we can use representations in a Hamel basis. But for a triangle, instead of a pentagon, the configuration $2,-1,-1$ leads uniquely to $1/2,1/2,-1$, then $1/2,-1/4,-1/4$, which is the homothetic by $1/4$ of the initial one, so the process never ends. One would need to find such a "repeating" model for the pentagon, in order to infirm the thesis.

whats this hamel basis?

If we consider the pentagon with $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},-(\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7})$. Supose that after $n$ steps each vertex have a $0$. Then $0=a_1\sqrt{2}+a_2\sqrt{3}+a_3\sqrt{5}+a_4\sqrt{7}-a_5(\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7})$ where $a_1+a_2+a_3+a_4+a_5=1$. In particular, each $a_i=1/2^{(b_i)}$ for a natural integer $b_i$. Then for some $k$ big enough, $2^ka^i$ is an integer for every i. Let $c_i=2^k(a_i-a_5)$ for $i=1,2,3,4$. Then $0=c_1\sqrt{2}+c_2\sqrt{3}+c_3\sqrt{5}+c_4\sqrt{7}$ for integers $c_i$. Then it is obvious that $c_i$ must be $0$. (just square a couple of times) But then $a_i-a_5=0$, so $a_i=a_5$ so $1=a_1+a_2+a_3+a_4+a_5=5a_5$. So $a_5=1/2^{(b_5)}=1/5$ and that is a contradiction.

Let $S$ be the set of rational numbers that can be expressed as $\displaystyle \frac{a}{2^b} - \frac{1}{5}$ with $a$ an integer and $b$ a non-negative integer. Then $\displaystyle \frac{-1}{5} \in S$, $\displaystyle \frac{4}{5} \in S$, $\displaystyle 0 \notin S$. Also note that $\displaystyle x, y \in S \Rightarrow \frac{x + y}{2} \in S$, which means if all the verticies of the pentagon start out with an element of $S$, they will forever have an element of $S$. So if we start by assigning four verticies the value $\displaystyle \frac{-1}{5}$ and one $\displaystyle \frac{4}{5}$, the value zero will never appear.

let us define $\frac{a}{b} \pmod{5}$ as $ab^{-1}$ then we place five numbers $1,1,1,1,-4$ then after any number of operations,the five numbers will be all congruent to $1$ mod $5$,contradiction.

If we start with any pentagon that has 3 or 5 non-zero numbers in it, you'll never get 5 zeroes. The reason for this is, that the only way to add zeroes replaces 2 non-zero numbers, $x$ and $-x$, with 2 zeroes. Therefore, the parity of the amount of zeroes never changes, and as it started even, it can never become 5, so it's impossible in those situations. Note: This means that even if you could choose any 2 numbers of the 5, ignoring adjacency, you'd get the same result.