The answer is ｢YES｣ Let p(1)＜...＜p(1990) are prime numbers and S=p(1)+...＋p(1990).Consider the 1990 numbers S!+p(1),...,S!+p(1990).Let's proof that these numbers are satisfied the conditions. ①pairwise coprime If ∃a（≧2） s.t. a|S!+p(i) , a|S!+p(j) (i＜j).a|p(j)-p(i) so a＜p(j).Then a|S! and so a|p(i),p(j).This is contradiction. ②sum of some numbers is composite Let consider k（≧2） numbers S!+p(i)...S!+p(j).Sum of these numbers is k･S!+p(i)+...＋p(j).This sum is divisible by 2≦a=p(i)+...＋p(j)≦S. The proof is completed.

The answer is yes. Let the prime numbers which are less than $1990$ be $2=p_1 < 3=p_2 < ... < p_n$. Let $a_i \equiv 1\pmod{p_j}$ for all $1\leq i \leq 1990$ and $1\leq j \leq n$. We know such $a_i$ 's exist because of the Chinese Remainder Theorem. So $a_i \equiv 1\pmod{p_1p_2...p_n}$ for all $1\leq i \leq 1990$. And also it is clear that we can select the $a_i$ 's such that they are all coprime and this sequence satisfies the condition above.