, 2010

We first construct the example for $n = 6$: $3, -5, 3, 3, -5, 3$.

Suppose that there exist $n \ge 7$ real numbers, that satisfy the conditions of the problem and choose any 5 consecutive numbers $a, b, c, d, e$. Using the conditions we get: $a+b+c > 0$, $c+d+e > 0 \Rightarrow (a+b+c+d+e) + c > 0$ and $a+b+c+d+e < 0$. This implies that $c > 0$, therefore for any 5 consecutive numbers one, which is in the middle, is always positive. From the last observation, it follows that all numbers except last two from both ends are positive.

Let us now consider 6 consecutive numbers: $a, b, c, d, e, f$. Using again our given conditions we get: $a+b+c+d+e < 0$ and $(a+b+c)+(d+e+f) > 0$. Thus, $f > 0$. By analogy, we have $a > 0$. This implies that all $n$ numbers must be positive and we get a contradiction.