Given distinct prime numbers $p$ and $q$ and a natural number $n \geq 3$, we aim to find all $a \in \mathbb{Z}$ such that the polynomial $f(x) = x^n + ax^{n-1} + pq$ can be factored into two integral polynomials of degree at least 1.

To solve this, we use the following reasoning:

1. Lemma (Eisenstein's Criterion Extension): If a polynomial $A(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ with integer coefficients is reducible in $\mathbb{Z}[x]$, and a prime $p$ divides $a_0, a_1, \ldots, a_{n-2}$ but does not divide $a_n$, and $p^2$ does not divide $a_0$, then $p$ does not divide $a_{n-1}$, and the polynomial $A(x)$ must have a rational root.

2. Applying this lemma to $f(x) = x^n + ax^{n-1} + pq$ with $p$ as a prime, if $f(x)$ is reducible, it must have a rational root. Since $f(x)$ is monic with integer coefficients, this rational root must be an integer.

3. Let $r$ be an integer root of $f(x)$. Then:
$$r^n + ar^{n-1} + pq = 0 \implies pq = -r^n - ar^{n-1}.$$
Since $pq$ is squarefree, $r$ must be $\pm 1$.

4. If $r = 1$:
$$1^n + a \cdot 1^{n-1} + pq = 0 \implies 1 + a + pq = 0 \implies a = -1 - pq.$$

5. If $r = -1$:
$$(-1)^n + a \cdot (-1)^{n-1} + pq = 0 \implies (-1)^n + (-1)^{n-1} a + pq = 0.$$
- If $n$ is odd:
$$-1 - a + pq = 0 \implies a = 1 + pq.$$
- If $n$ is even:
$$1 + a + pq = 0 \implies a = -1 - pq.$$

Thus, the values of $a$ that allow the polynomial $f(x) = x^n + ax^{n-1} + pq$ to be factored into two integral polynomials of degree at least 1 are:
$$a = -1 - pq \quad \text{and} \quad a = 1 + pq \quad \text{(if \( n \) is odd)}.$$

The answer is: $\boxed{-1 - pq \text{ and } 1 + pq}$.