Let $S$ be a set with $|S| = 35$. A set $F$ of mappings from $S$ to itself is said to satisfy property $P(k)$ if for any $x, y \in S$, there exist $f_1, f_2, \ldots, f_k \in F$ (not necessarily different) such that $f_k(f_{k-1}(\cdots (f_1(x)) \cdots )) = f_k(f_{k-1}(\cdots (f_1(y)) \cdots ))$.

We aim to find the least positive integer $m$ such that if $F$ satisfies property $P(2019)$, then it also satisfies property $P(m)$.

To determine this, consider a minimal length sequence of mappings $f_k, f_{k-1}, \ldots, f_1$ such that $f_k(f_{k-1}(\cdots (f_1(x)) \cdots )) = f_k(f_{k-1}(\cdots (f_1(y)) \cdots ))$ for fixed $x, y \in S$. Denote $g_i(x) = f_i(f_{i-1}(\cdots (f_1(x)) \cdots ))$, with $g_0(x) = x$.

Let $A_i$ be the unordered pair $(g_i(x), g_i(y))$. The key claim is that $A_0, A_1, \ldots, A_k$ are all distinct, and $A_k$ is the only pair consisting of two equal elements. If there exist two equal pairs $A_i$ and $A_j$ (where $i < j$), we can use the functions $f_k, f_{k-1}, \ldots, f_{j+1}, f_i, f_{i-1}, \ldots, f_1$ instead to obtain equal final values, contradicting the assumption that $f_k, f_{k-1}, \ldots, f_1$ is a minimal length sequence.

Hence, the maximum length of the sequence is at most the number of unordered pairs of distinct elements, which is exactly $\binom{35}{2}$.

To construct such a sequence, let $S = \{0, 1, \ldots, 34\}$ and define two mappings $f(x)$ and $g(x)$ as follows:
$$f(x) = (x + 1) \pmod{35},$$
$$g(0) = 1, \quad g(x) = x \text{ for all } 1 \leq x \leq 34.$$

Using these functions on $(x, y) = (1, 18)$, we apply $f$ 34 times to turn $(1, 18)$ into $(0, 17)$, then apply $g$ to turn it into $(1, 17)$. Repeating this process another 16 times yields $(1, 1)$ after $35 \times 17 = 595 = \binom{35}{2}$ functions.

Thus, the least positive integer $m$ such that if $F$ satisfies property $P(2019)$, then it also satisfies property $P(m)$ is $\binom{35}{2}$.

The answer is: $\boxed{595}$.