All-Soviet-Union Mathematical Olympiad

Solution:
Answer $1989$

Since $|a - b| \leq \max(a, b)$, a trivial induction shows that the expression does not exceed $\max(a_1, a_2, \ldots, a_{1990}) = 1990$. But for integers, $|a - b|$ has the same parity as $a + b$, so a trivial induction shows that the expression has the same parity as $a_1 + a_2 + \ldots + a_{1990} = 1990 \cdot 1991 / 2$, which is odd. So it cannot exceed $1989$. That can be attained by the permutation $2, 4, 5, 3, 6, 8, 9, 7, \ldots, 4k + 2, 4k + 4, 4k + 5, 4k + 3, \ldots, 1984 + 2, 1984 + 4, 1984 + 5, 1984 + 3, 1990, 1$. Because we get successively $2, 3, 0; 6, 2, 7, 0; 10, 2, 11, 0; \ldots; 4k + 2, 2, 4k + 3, 0; \ldots; 1986, 2, 1987, 0; 1990, 1989$.