Find all functions $f : Q \to Q$, such that $$f(x + f (y + f(z))) = y + f(x + z)$$ for all $x ,y ,z \in Q$ . 

(I. Voronovich)
1. **Define the assertion**: Let \( P(x, y, z) \) be the assertion of the given functional equation:
   \[
   f(x + f(y + f(z))) = y + f(x + z)
   \]
   for all \( x, y, z \in \mathbb{Q} \).

2. **Initial simplification**: Let \( a = f(0) \). Consider the assertion \( P(0, y, 0) \):
   \[
   f(f(y + f(0))) = y + f(0)
   \]
   Substituting \( f(0) = a \), we get:
   \[
   f(f(y + a)) = y + a
   \]
   This implies that \( f \) is surjective.

3. **Involution property**: To show that \( f \) is an involution, consider \( P(0, y - a, 0) \):
   \[
   f(f(y)) = y
   \]
   This shows that \( f \) is an involution, i.e., \( f(f(x)) = x \) for all \( x \in \mathbb{Q} \).

4. **New assertion**: Applying \( f \) to both sides of the original equation \( P(x, y, z) \), we get:
   \[
   f(f(x + f(y + f(z)))) = f(y + f(x + z))
   \]
   Using the involution property \( f(f(t)) = t \), we simplify this to:
   \[
   x + f(y + f(z)) = f(y + f(x + z))
   \]
   Let this be the new assertion \( Q(x, y, z) \).

5. **Simplification using \( Q \)**: Consider \( Q(f(y) - a, x, a) \):
   \[
   f(x + y) = f(x) + f(y) - a
   \]
   This shows that \( f \) is a Cauchy additive function with a constant shift \( a \).

6. **Form of \( f \)**: Since \( f \) is defined on \( \mathbb{Q} \), which is a vector space over \( \mathbb{Q} \), and \( f \) is additive, it must be of the form:
   \[
   f(x) = cx + a
   \]
   for some constants \( c \) and \( a \).

7. **Substitute back**: Substitute \( f(x) = cx + a \) into the original equation:
   \[
   f(x + f(y + f(z))) = y + f(x + z)
   \]
   This becomes:
   \[
   c(x + c(y + cz + a) + a) + a = y + c(x + z) + a
   \]
   Simplifying, we get:
   \[
   c(x + cy + c^2z + ca + a) + a = y + cx + cz + a
   \]
   Equating coefficients, we find:
   \[
   c^2 = 1 \quad \text{and} \quad ca + a = 0
   \]
   Thus, \( c = 1 \) or \( c = -1 \). If \( c = 1 \), then \( a = 0 \). If \( c = -1 \), then \( a \) can be any rational number.

8. **Conclusion**: The solutions are:
   \[
   \boxed{f(x) = x \quad \forall x \in \mathbb{Q}}
   \]
   and
   \[
   \boxed{f(x) = a - x \quad \forall x \in \mathbb{Q}, \text{ where } a \in \mathbb{Q}}
   \]