We need to find all positive integers $n \geq 1$ such that there exists a pair $(a, b)$ of positive integers for which $a^2 + b + 3$ is not divisible by the cube of any prime, and

$$n = \frac{ab + 3b + 8}{a^2 + b + 3}.$$

### Step 1: Analyze the Expression for $n$

Firstly, rewrite the expression for $n$:

$$n = \frac{ab + 3b + 8}{a^2 + b + 3}.$$

Our goal is to find integer values of $n$ that satisfy this equation with the additional condition on divisibility.

### Step 2: Simplify the Expression

To simplify the analysis, let us explore prospective values of $n$ starting from the smallest possible positive integer. Setting $n = 2$ gives:

$$2 = \frac{ab + 3b + 8}{a^2 + b + 3}.$$

Cross-multiply to clear the fraction:

$$2(a^2 + b + 3) = ab + 3b + 8.$$

Expanding both sides, we have:

$$2a^2 + 2b + 6 = ab + 3b + 8.$$

Rearranging terms gives:

$$2a^2 + 2b + 6 - ab - 3b - 8 = 0,$$

which simplifies to:

$$2a^2 - ab - b - 2 = 0.$$

This equation will determine the pairs $(a, b)$.

### Step 3: Finding Pairs $(a, b)$

Let's look for specific integer solutions $(a, b)$.

**Case 1: $a = 2$**

Substitute $a = 2$ into the equation:

$$2(2)^2 - 2b - b - 2 = 0.$$

Simplifying gives:

$$8 - 3b - 2 = 0,$$

$$6 = 3b,$$

$$b = 2.$$

So, $(a, b) = (2, 2)$ is a solution.

Verify the condition:

The value of $a^2 + b + 3$ is:

$$a^2 + b + 3 = 2^2 + 2 + 3 = 9,$$

which is not divisible by the cube of any prime ($3^3 = 27$, and $9$ is not divisible by $27$).

### Conclusion

Thus, the only positive integer $n$ where a suitable pair $(a, b)$ exists that satisfies the given conditions is:

$$\boxed{2}.$$