CMO 1988 Problem 3

Given a finite sequence of real numbers $a_1,a_2,\dots ,a_n$ ( $\ast$ ), we call a segment $a_k,\dots ,a_{k+l-1}$ of the sequence ( $\ast$ ) a "long"(Chinese dragon) and $a_k$ "head" of the "long" if the arithmetic mean of $a_k,\dots ,a_{k+l-1}$ is greater than $1988$ . (especially if a single item $a_m>1988$ , we still regard $a_m$ as a "long"). Suppose that there is at least one "long" among the sequence ( $\ast$ ), show that the arithmetic mean of all those items of sequence ( $\ast$ ) that could be "head" of a certain "long" individually is greater than $1988$ .