Let
$$f(x) = (x+1)(x+2)\cdots(x+n)+2. \qquad \textcircled{1}$$
Obviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We are going to prove that $f(x)$ has property (2).
For any integer $t$, since $n \ge 4$, we know that there exists definitely a multiple of 4 in any $n$ consecutive numbers $t+1, t+2, \dots, t+n$. Then from $\textcircled{1}$, we have $f(t) \equiv 2 \pmod 4$.
Then for any $k$ ($k \ge 2$) positive integers $r_1, r_2, \dots, r_k$, we have
$$f(r_1)f(r_2)\cdots f(r_k) \equiv 2^k \equiv 0 \pmod 4.$$
On the other hand, for any positive integer $m$, we have
$f(m) \equiv 2 \pmod{4}$. Therefore,
$$f(m) \not\equiv f(r_1)f(r_2)\cdots f(r_k) \pmod{4},$$
which implies that $f(m) \neq f(r_1)f(r_2)\cdots f(r_k)$. We then find the required $f(x)$ and complete the proof. $\square$