Solution:
The answer is that all the numbers must be zero. (Clearly, this works.)

We now prove this is the only solution. Call the numbers $x_{1}, x_{2}, \ldots, x_{100}$. Then the sum $x_{1}+x_{2}+x_{3}$ must be even, since it is either $2x_{1}$, $2x_{2}$, or $2x_{3}$. Similarly, $x_{2}+x_{3}+x_{4}$ must be even.

In this way, $x_{1}$ and $x_{4}$ have the same parity. By the same reasoning, $x_{4}$ and $x_{7}$ have the same parity, and so on—the numbers $x_{k}$ and $x_{k+3}$ have the same parity. Since $3$ doesn't divide $100$, that means all the numbers have the same parity. Clearly then all the numbers are even (rather than all odd).

We may now employ infinite descent: if $(x_{1}/2, \ldots, x_{100}/2)$ is a working assignment, then so is $(x_{1}/2, x_{2}/2, \ldots, x_{100}/2)$, and then so is $(x_{1}/4, x_{2}/4, \ldots, x_{100}/4)$. Such a process cannot go on indefinitely unless $x_{k}=0$ for all $k$, completing the proof.