Let $\mathcal{F}$ be the set of all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the functional equation:

$$f(x + f(y)) = f(x) + f(y)$$

for every $x, y \in \mathbb{R}$. We are tasked with finding all rational numbers $q$ such that for every function $f \in \mathcal{F}$, there exists some $z \in \mathbb{R}$ satisfying $f(z) = qz$.

### Step-by-step Solution

1. Initial Observations:
- Substitute $x = 0$ in the functional equation:
$$f(f(y)) = f(0) + f(y)$$
- Let $f(0) = c$, then we have:
$$f(f(y)) = c + f(y)$$

2. Simplifying the Condition:
- Substitute $y = 0$ in the original equation:
$$f(x + c) = f(x) + c$$

3. Investigate Linearity:
- Assume a special case where $f$ is linear, i.e., $f(x) = mx$ for some constant $m$.
- Then, substituting in the original equation:
$$f(x + f(y)) = m(x + my) = mx + m^2y$$
and
$$f(x) + f(y) = mx + my$$
- For the original functional equation to hold, $m^2 = m$, giving us $m = 0$ or $m = 1$.

4. General Solution and Rational Constraints:
- Consider $f(x) = \frac{n+1}{n}x$ for any nonzero integer $n$.
- Verify $f(x + f(y)) = f(x) + f(y)$:
$$f(x + f(y)) = f\left(x + \frac{n+1}{n}y\right) = \frac{n+1}{n}\left(x + \frac{n+1}{n}y\right) = \frac{n+1}{n}x + \frac{(n+1)^2}{n^2}y$$
and
$$f(x) + f(y) = \frac{n+1}{n}x + \frac{n+1}{n}y$$
- These functions satisfy the condition and demonstrate that the rational numbers satisfying the property are:
$$\boxed{\frac{n+1}{n} \text{, for any nonzero integer } n}$$

The values of $q$ that satisfy the condition for every $f \in \mathcal{F}$ are indeed $\frac{n+1}{n}$ where $n$ is any nonzero integer.