The answer is $\boxed{16}$. Proof $P$ denotes the set of all prime numbers. Since $\#\left\{p^{2} |p\in \mathbb{P} ,p^{2} < 2018\right\} \cup \{1\} =15$, we must have $k >15$. Suppose we are given a set $S$ of $16$ distinct and pairwise coprime naturals smaller than $2018$. Assume that they are all composite numbers except for $1$ (if $1\in S$). For each $n\in \mathbb{N}^{*}$, let $f( n)$ be the smallest prime divisor of $n$ and $f( 1) =1$. If $n$ is a composite numer and $n< 2018$, then $f( n) \leq 43$. So, for each $n\in S$, $f( n) \leq 43$, of course. Since $\#\{p\in \mathbb{P} |p\leq 43\} \cup \{1\} =15$, by the pigeonhole principle, there exists $( x,y) \in S\times S$ such that $f( x) =f( y)$ and $x\neq y$. This clearly contradicts that the given $16$ naturals are pairwise coprime. Therefore, $S$ must contain at least one prime number. $\blacksquare$