Let $\mathbb{Q}$ be the set of rational numbers. We have a function $f: \mathbb{Q} \to \mathbb{Q}$ that satisfies the property such that for every $x, y \in \mathbb{Q}$:

$$f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y).$$

Our task is to show that there exists an integer $c$ such that for any aquaesulian function $f$, there are at most $c$ different rational numbers of the form $f(r) + f(-r)$ for some rational number $r$, and to find the smallest possible value of $c$.

### Solution

1. **Analyzing the conditions of $f$**:

Consider the property $f(x+f(y)) = f(x) + y$. If we set $x = 0$, we get:

$$f(f(y)) = f(0) + y.$$

This implies that the function $f$ behaves as a partial inverse.

Similarly, if we consider the condition $f(f(x)+y) = x + f(y)$ and set $y = 0$, we get:

$$f(f(x)) = x + f(0).$$

This suggests involution or a linear shift in behavior of the function.

2. **Deducing the expression for $f(r) + f(-r)$**:

With both transformations $f(x+f(y)) = f(x) + y$ and $f(f(x)+y) = x + f(y)$ leading to similar structures, let's see what happens when we consider $r$ and $-r$:

If we apply $f$ on $f(r) + f(-r)$, using the equations above and assuming one transformation, say:

$$f(f(r) + f(-r)) = r + f(-r).$$

and

$$f(f(-r) + f(r)) = -r + f(r).$$

Adding these derived results can lead to:

$$f(f(r)) + f(f(-r)) = (f(0) + r) + (f(0) - r) = 2f(0).$$

Hence, $f(r) + f(-r)$ always simplifies to $2f(0)$.

Therefore, regardless of $r$, $f(r) + f(-r)$ resolves to a constant.

3. Conclusion:

Since each $f(r) + f(-r)$ results in a single fixed rational number $2f(0)$, the integer $c$ representing the maximum number of distinct values for $f(r) + f(-r)$ must be:

$$\boxed{1}.$$

This tells us that the smallest possible value of $c$ ensuring the uniqueness condition for any aquaesulian function is indeed 1.