Is there anybody who could post all of the problems of this contest?

$n=5$ is the minimum value that satisfies the condition Indeed, let $n=4$. By a) and b), it's easy to see that $a_1>0>a_4$. We will investigate some cases Case 1: $a_1\geq a_2>0>a_3>a_4$. We can set $a_1=a,a_2=b,a_3=-c,a_4=-d$. Then \[ \begin{cases} a+b>c+d\\ a^3+b^3<c^3+d^3\\ a^5+b^5>c^5+d^5 \end{cases} \]Since $a\geq b>0, c\geq d>0$, hence from the system above, we have $a>d,c>b$. On the other hand, by $a^3+b^3-c^3<d^3<(a+b-c)^3$, we deduce \[ 3(a+b)(a-c)(b-c)>0 \]Great, now we have $c>a\geq b$. From here, with AM-GM, we can easily have \[ a^5+b^5=a^3(a^2-b^2)+b^2(a^3+b^3)<c^3(a^2-b^2)+b^2(c^3+d^3)\leq\frac{2a^5+2b^5+3c^5+3d^5}{5} \]\[ \Rightarrow a^5+b^5<c^5+d^5\quad\text{(contradict to c))} \]Case 2: $a_3\leq a_2\leq 0$. We derive $a_1>-(a_2+a_3+a_4)$ from a). Hence \[ a_1^3+a_2^3+a_3^3+a_4^3>-3(a_2+a_3)(a_2+a_4)(a_3+a_4)\leq 0\quad\text{(contradicts with b))} \]Case 3: $a_2\geq a_3\geq 0$. From b) we get $a_1^3+a_2^3+a_3^3<-a_4^3$. Thus $0\leq a_3\leq a_2\leq a_1<-a_4$. So \[ a_1^5+a_2^5+a_3^5+a_4^5<a_4^2(a_1^3+a_2^3+a_3^3+a_4^3)<0\quad\text{(contradicts with c))} \]With $n<4$, we can add $4-n$ zeroes, so we will have a $4$-tuple, and there are no tuple satisfies the given condition, as shown above For $n=5$, we can find $-7,-7,2,5,8$ satisfies. I guess we can find a lot more with Maple