In the question replace $1985 \to n$, and $662 \to \frac{n}{3}$, and use induction. The statement is trivially true for $n=1,2,3$, now suppose it is true for $n\le k$ and consider $n=k+3$. Among the $k+3$ points find some sequence that goes $+1, \underbrace{-1,-1,\dots,-1}_{\ge 1},+1$. Now remove the two $+1$'s and one $-1$. This leaves $n$ points, so by our assumption there is a good point, $p$ (which must be poisitive). Now when we replace the points we removed, the point $p$ is still a good point because the two positive values lie either side of the negative value and therefore don't decrease the partial sums from either direction.

Just another solution: We use n and n/3 again.. Assume towards a contradiction that there exists no good point. Then, for every 1, consider the contiguous sum of smallest length that starts with that 1 and equals 0. Clearly, that sum ends with a -1. Now, every 1 corresponds to exactly 1 -1 (if there are 2 sequences of minimal length, pick the one that goes right). For every -1, there can be at most 2 1s that correspond to it (this isn't hard to see.) But this is a contradiction since if we have <n/3 ones, we can't possibly have this correspondence.

riemannHypothesis2.71828 wrote: For every -1, there can be at most 2 1s that correspond to it (this isn't hard to see.) $1$ on the $B,C,D$ correspond to $-1$ on $A$.

Attachments:

Consider three consecutive blocks of equal lengths such that the first and third blocks consist of +1s and the middle block consists of -1s. Remove such blocks repeatedly until no -1 remains (clearly possible since # of +1s > twice # of -1s). At the end, the remaining +1 is good. Obviously it is also good in the beginning.

Yinxiu, what drawing program did you use? Thanks.

The Geometer'S Sketchpad