WLOG n=3 and we can consider it as an old problem.

note that if $S_n$ is the set of functions that satisfies $\sum_{i=1}^{n}(a_i-a_{i+1})f(a_i+a_{i+1}) = 0$ then $S_{n} \subseteq S_{n-1}$ so $S_n \subseteq S_3$ so it suffices to consider $n=3$ $$(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$$let $P(a,b,c)$ is the previous assertion $P(a,a+d,a+2d) \implies f(2a+d)+f(2a+3d)=2f(2a+2d)$ so $f(x+y)=2f(\frac{x+y}{2})$ so its jensen and so $f(x)=kx \forall x \in \mathbb{Q}$ combining with $(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$ $$f(a)(c-b)+f(b)(a-c)+f(c)(b-a)=0$$put $a,b \in \mathbb{Q} \implies f(a)=ka \forall a \in \mathbb{R}^+$ and we win

Ali3085 wrote: note that if $S_n$ is the set of functions that satisfies $\sum_{i=1}^{n}(a_i-a_{i+1})f(a_i+a_{i+1}) = 0$ then $S_{n} \subseteq S_{n-1}$ so $S_n \subseteq S_3$ so it suffices to consider $n=3$ $$(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$$let $P(a,b,c)$ is the previous assertion $P(a,a+d,a+2d) \implies f(2a+d)+f(2a+3d)=2f(2a+2d)$ so $f(x+y)=2f(\frac{x+y}{2})$ so its jensen and so $f(x)=kx \forall x \in \mathbb{Q}$ combining with $(a-b)f(a+b)+(b-c)f(b+c)+(c-a)f(a+c)=0$ $$f(a)(c-b)+f(b)(a-c)+f(c)(b-a)=0$$put $a,b \in \mathbb{Q} \implies f(a)=ka \forall a \in \mathbb{R}^+$ and we win Watch carefully on your Jenson equation lol. Cuz the true solution is all linear functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$