Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ 
Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ such that
$$\left\{\begin{array}{l} a_{1}+a_{2}+\ldots+a_{n}>0 \\a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}<0 \\a_{1}^{5}+a_{2}^{5}+\ldots+a_{n}^{5}>0 \end{array}\right.$$

I don't have a solution, but I suspect $n_{\min}=5$ is optimal. I think so mainly because I could not find anything by brute force for either $4$ or $3$. There are quite a few solutions for $n=5$, for example:
$$a_{1}=1, a_{2}=-0.88, a_{3}=-0.88, a_{4}=0.5, a_{5}=0.5$$
$$a_{1}=-8, a_{2}=-7, a_{3}=3, a_{4}=4, a_{5}=9$$
It is also not difficult to prove by contradiction that $n_{\min}>2$.

n = 3 seems to be fine. Let $c$ be a parameter to be determined later.
Put $a_1 =-1-6c^3$, $a_2=-1+6c^3$, and $a_3=6c^2$.
Clearly $a_1 + a_2 + a_3 = -1 +6c^2$ which is positive for $c$ large enough.
Now, $a_1^3 + a_2^3 +a_3^3 = (-1-6c^3)^3 + (-1 + 6c^3)^3 + (6c^2)^3 = -2 < 0 $
Also, $a_1^5 + a_2^5 + a_3^3 = (-1 - 6c^3)^5 + (-1 + 6c^3)^5 + (6c^2)^5$ has the leading coefficient $2*6^4*c^{12}$ that is it, it is positive for $c$ large enough.
