We are asked to find all functions $f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that for all rational numbers $x < y < z < t$ that form an arithmetic progression, the following relation holds:
$$f(x) + f(t) = f(y) + f(z).$$
To begin, consider that an arithmetic progression with four terms $x, y, z, t$ must have a common difference, say $d$. Therefore, we can express the terms as:
$$y = x + d, \quad z = x + 2d, \quad t = x + 3d.$$
Substituting these into the functional equation, we have:
$$f(x) + f(x + 3d) = f(x + d) + f(x + 2d).$$
This is a functional equation often satisfied by linear functions. To verify, assume $f(x) = Ax + B$ for some $A, B \in \mathbb{Q}$. Then:
$$f(x) = Ax + B, \quad f(x + d) = A(x + d) + B = Ax + Ad + B,$$
$$f(x + 2d) = A(x + 2d) + B = Ax + 2Ad + B, \quad f(x + 3d) = A(x + 3d) + B = Ax + 3Ad + B.$$
Substituting into the functional equation, we check:
$$(Ax + B) + (Ax + 3Ad + B) = (Ax + Ad + B) + (Ax + 2Ad + B).$$
Simplifying both sides, we get:
$$2Ax + 3Ad + 2B = 2Ax + 3Ad + 2B.$$
The equation holds true, confirming that any linear function of the form $f(x) = Ax + B$ satisfies the given condition.

To ensure this is the only solution, consider the nature of an arithmetic progression and the structure of linear functions. The symmetry and structure imposed by the condition, combined with the density of rational numbers, indicate that deviations from linearity would involve adding non-linear terms that disrupt the equality for all rational progressions.

Hence, the solution is:
$$f(x) = Ax + B \text{ for some } A, B \in \mathbb{Q}.$$

Finally, the form of all such functions is:
$$\boxed{f(x) = Ax + B, \text{ where } A, B \in \mathbb{Q}}.$$