A bit strange solution: Assume that the set $\{a_1,...,a_m\}$ has the desired property. Then by Cauchy-Schwarz: $$6=\sum_{i=1}^na_i^3a_i^2\leq\sqrt{\sum_{i=1}^na_i^6\sum_{i=1}^na_i^4}=\sqrt{35}$$a contradiction.

I don't think the Cauchy-Schwarz solution is strange at all, it was really the first thing that came to my mind when I saw that problem. Of course you could also argue with pretty much any other triple and get nicer numbers, such as \[21=\left(\sum x_i^2\right)\left(\sum x_i^6\right) \ge \left(\sum x_i^4\right)^2=25.\]This problem turned out to be surprisingly difficult (in terms of the average score). The main difficulty seems to be to even think of inequalities here. But if you have seen similar problems such as Problem 5 from the IMO 1979, this is really very straightforward.

\[15=\left(\sum x_i^2\right)\left(\sum x_i^4\right) \ge \left(\sum x_i^3\right)^2=16.\]

Lemma:$(a_1^2 + a_2^2 + ... + a_n^2)(a_1^6+a_2^6 + ... + a_n^6) \ge (a_1^4 + a_2^4 +... +a_n^4)^2$ $$3 \cdot 7 \ge 25$$Contradiction

Suppose so. Let the real numbers be numbered $a_1,a_2,\ldots,a_n$ and notice that by C-S: $$\left(\sum_{i=1}^na_i^4\right)\left(\sum_{i=1}^na_i^8\right)\ge\left(\sum_{i=1}^na_i^6\right)^2$$$$\Leftrightarrow5\cdot9\ge7^2$$which is clearly false. Hence no such set can exist.