We are tasked with finding all pairs of integers $(a, b)$ such that there exist functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ and $g: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying the conditions:

$$f(g(x)) = x + a \quad \text{and} \quad g(f(x)) = x + b$$

for all integers $x$.

To solve this problem, we will analyze the functional equations given and deduce the necessary conditions for $(a, b)$.

1. Analyzing the Equations:

From the first equation, $f(g(x)) = x + a$, applying $g$ on both sides, we get:
$$g(f(g(x))) = g(x + a).$$

Using the second equation $g(f(x)) = x + b$, substitute $y = g(x)$, we have:
$$f(y) = g^{-1}(y - b).$$

Applying $f$ on both sides of the equation $g(f(x)) = x + b$, we have:
$$f(g(f(x))) = f(x + b).$$

2. Substitution and Inferences:

Use the result from applying $g$ on the first equation:
$$g(f(g(x))) = x + a + b.$$

Since we also have $g(f(g(x))) = g(x + a)$, equating both gives:
$$g(x + a) = x + a + b.$$

$$\Rightarrow g(x) = x + b - a.$$

From this, we realize that $g$ is a linear function. Substitute $x = g(y)$ into $f(g(x)) = x + a$:
$$f(x + b - a) = x + a.$$

3. Consistency Check:

From these conditions, we find that both $g$ and $f$ imply a consistent cyclic nature where:
$$f(y + b - a) = y + a \quad \Rightarrow \quad y + b - a + a = y + b.$$

Therefore, the cycle completes without contradiction if the magnitude of the shift imposed by $a$ equals that by $b$, suggesting:
$$|a| = |b|.$$

Conclusion:

Through the analysis of the problem's functional equations and the cycle of applications between $f$ and $g$, the condition:

$$\left| a \right| = \left| b \right|$$

is necessary and sufficient for the functions $f$ and $g$ to exist satisfying the given conditions for the pair $(a, b)$. Therefore, the solution is $\boxed{\left| a \right| = \left| b \right|}$.