Assume Goldbach's Conjecture. Then choose the first $10^{2002} + 1$ primes. Their product $P$ will be even, and so will their sum $S < P$. Thus $P-S$ is even and can be written as $p+q$ with them both prime. Adding $p, q$ to our list of primes, we have $S' = S + p + q = P$, and $P' = Ppq$ which is divisible by $S'$.

Choose $10^{2002}$ arbitrary primes. Let the product be $P$, and the sum be $S$. Consider the smallest prime factor of $P-S$, $q$. Then $P-S$ can be written as $\sum_{n=1}^{N}q$. Adding $q, \ldots, q$ to our list of primes, we have $S' = S + q + \ldots + q = P$, and $P' = q^{N} \cdot P$ which is divisible by $S'$.