For a given positive integer $n$ and prime number $p$, we aim to find the minimum value of the positive integer $m$ that satisfies the following property: for any polynomial
$$f(x) = (x + a_1)(x + a_2) \ldots (x + a_n)$$
where $a_1, a_2, \ldots, a_n$ are positive integers, and for any non-negative integer $k$, there exists a non-negative integer $k'$ such that
$$v_p(f(k)) < v_p(f(k')) \leq v_p(f(k)) + m.$$
Here, $v_p(N)$ denotes the largest non-negative integer $t$ such that $p^t \mid N$ for a non-zero integer $N$.

The minimum value of $m$ that satisfies this property is:
$$n + v_p(n!).$$

The answer is: \boxed{n + v_p(n!)}.