Positively non-positive

We will show that only $N=3,4$ work. $$ Consider the sequence $$ $(N-3) + 1 + \underbrace{ {(-1) + (-1)..+ (-1)} }_{ \text{ N-2 times } } = 0$ $$ These are $N$ real numbers whose sum is equal to $0$ . $$ Now we consider their second summation. That equals $$

$(N-3)*1 - 1 + \underbrace{ {1+1+1..+1 } }_{ \text{N-4 times } } - (N-3) = (N-4)$ which is positive when $N \geq 5$ This contradicts the second inequality. So all $N \geq 5$ don't always work. $$

Now consider the cases when $N=3, N=4$ $$ Case 1 : When $N=3$ : Let the numbers be $a_1 , a_2 , a_3$
$a_1 + a_2 +a_3 = 0$ so , $a_1 + a_2 = -a_3$ It can easily be checked that the second summation in this case is always $\leq 0$ $$

Case 2 : When $N=4$ : Let the numbers be $a_1 , a_2 , a_3 , a_4$ In this case the second summation after simplifications reduces to $-(a_2 + a_4)^2$ which is again non positive. $$ So we conclude that only $N=3$ and $N=4$ work, and hence the answer is $\boxed{2}$

If $N \ge 5$ , we can choose $a_1 = -1, a_N = -1, a_3 = 2, a_2 = a_4 = ... = a_{N - 1} = 0$ , which contradicts the second inequality and satisfy the first equality.

If $N = 4$ , note that the second inequality can be written as $(a + b)(c + d) = -(a + b)^2 \le 0$ .

Finally, if $N = 3$ , $ab + ac + bc = c(a + b) + ab = -(a + b)^2 + ab$ .

We claim that $(a + b)^2 \ge ab$ .

Note that by $a + b + c = 0$ , at least one of $a, b, c$ is nonpositive and at least one is nonnegative.

If $bc \le 0$ , we're done, since squares are nonnegative.

If not, $bc > 0$ , the inequality follows from $(b - c)^2 \ge 0 \Rightarrow b^2 + c^2 \ge 4bc \ge bc$ .

Thus, there are $\boxed{2}$ possible values of $N$ .

Note : I think the problem should include the word inclusive to avoid ambiguity. I submitted $1$ the first time for not including $N = 3$ .

For all $N \geq 5$ , we consider the following construction:

• $a_1 = a_{N-1} = 0$
• $a_2 = a_3 = a_4 = ... = a_{N-2} = 1$
• $a_N = -(a_1 + a_2 + a_3 + ... a_{N-1}) = -(N-3)$

this includes the condition in the question

Now, with this construction, we consider the sum:

$a_1 a_2 + a_2 a_3 + ... + a_{N-1} a_N + a_N a_1$

$= a_2 a_3 + a_3 a_4 + a_4 a_5 + ... a_{N-3} a_{N-2}$

$= 1 + 1 + 1 + 1 + ... + 1$

$= N - 4$

$\geq 1 > 0$

Hence, for all $N \geq 5$ , the statement is false by construction. However, obviously, if $N < 5$ , this construction would not hold. Therefore, we will instead prove that the statement is true for $N = 3$ and $N = 4$ separately.

$N = 3:$

$a_1 a_2 + a_2 a_3 + a_3 a_1$

$= a_1 a_2 + (a_1 + a_2) ( - a_1 - a_2)$ , by the condition given

$= - a_1^2 - a_2^2 - a_1 a_2$

$= - (a_1 + \frac{1}{2} a_2)^2 - \frac{3}{4} a_2^2$

$\leq 0$ for all $a_1 , a_2$

$N = 4:$

$a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1$

$= a_1 a_2 + a_2 a_3 + (a_1 + a_3) ( - a_1 - a_2 - a_3)$ , by the condition given

$= - a_1^2 - a_3^2 - 2 a_1 a_3$

$= - (a_1 + a_3)^2$

$\leq 0$ for all $a_1 , a_2 , a_3$

Hence, the statement is only true for $N = 3 , 4$ , and the answer is $2$ .

Notice that $0 = (a_1 + \cdots + a_N)^2 = a_1^2 + \cdots a_N^2 + \sum_{i\neq j} a_i a_j .$ Let's split the last sum as $2T + V$ where $T = a_1a_2 + a_2a_3 + \cdots + a_N a_1$ and in $V$ we collect the remaining terms. $V$ contains $O(N^2)$ terms whereas $T$ only $N$ terms. Therefore we expect $T \leq 0$ to be generically false. To be explicit, for $N \geq 6$ even, the set $a_1 = \cdots = a_{N/2} = 1$ , $a_{N/2+1} = \cdots = a_N = -1$ gives positive $T$ . For $N \geq 5$ odd, take the same configuration on the first $N-1$ terms and $a_N = 0$ .

Let's turn to the small cases. For $N = 3$ the summand $V$ always vanishes and so $2T = -a_1^2 -a_2^2 - a_3^2 \leq 0.$ For $N = 4$ we get $2T = -(a_1 + a_3)^2 - (a_2 + a_4)^2 \leq 0.$ So $N = 3, 4$ are the only integers for which the statement is true and the answer is $\bf 2$ .

For $N=3$ , the expression $=a_1a_2+a_2a_3+a_3a_1=a_1a_2-(a_1+a_2)^2=-(a_1+\frac{a_2}{2})^2-\frac{3a_2^2}{4}<0$ . For $N=4$ , the expression =a_1a_2+a_2a_3+a_3a_4+a_4a_1=(a_1+a_3)(a_4+a_2)=-(a_1+a_3}^2 <0. For $N=5$ , let the numbers are $-9,-4,1,5,7$ . Then numerical value of the expression is $9$ . So for $N=5$ , the expression is positive. For $N=6$ , embed a $0$ in between $-4$ and $1$ . So the sequence is $-9,-4,0,1,5,7$ and the value of the expression is $13$ . Now, for each subsequent cases, embed a $0$ in between $0$ and $1$ , and for each cases we get a new sequence whose sum is $0$ and the value of the expression is $13$ . So answer is $2$ .

We first claim that if $N > 4$ , we can always choose a sequence $\{ a_i \}_{i=1}^N$ such that $\sum_{i=1}^N a_i = 0$ but $a_1 a_N + \sum_{i=1}^{N-1} a_i a_{i+1} > 0$ . Consider $a_i = 1$ for $i \in \{ 1, 2, \ldots, N-2 \}$ , $a_{N-1} = -1$ , hence $a_N = 3-N$ . Then $\begin{aligned} a_1 a_N + \sum_{i=1}^{N-1} a_i a_{i+1} &= 3-N + \sum_{i=1}^{N-3} 1 + a_{N-2} a_{N-1} + a_{N-1} a_N \\ &= (3-N) + (N-3) + (1)(-1) + (-1)(3-N) \\ &= N-4 > 0. \end{aligned}$ Now consider the case $N = 3$ . We have $a_3 = -(a_1 + a_2)$ and therefore $a_1 a_2 + a_2 a_3 + a_3 a_1 = -(a_1^2 + a_1 a_2 + a_2^2)$ . If $a_1 a_2 \ge 0$ , this expression is clearly nonpositive. If $a_1 a_2 < 0$ , then we write $-(a_1^2 + a_1 a_2 + a_2) = -((a_1 + a_2)^2 - a_1 a_2)$ and again it is obvious that this expression must be nonpositive. So the condition for $N = 3$ is always satisfied.

For the case $N = 4$ , with $a_4 = -(a_1 + a_2 + a_3)$ , algebraic simplification of the LHS condition easily gives $a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1 = (a_1 + a_3)(a_2 + a_4) = -(a_1 + a_3)^2,$ which is obviously nonpositive. This covers all cases; therefore, the answer is $\boxed{2}$ .

2

2