Taiwan IMO Selection Camp

Solution: First, setting $a_4 = a_5 = \cdots = a_n = a_1$ in the original equation, we get
$$\sum_{i=1}^{3} (a_i - a_{i+1}) f(a_i + a_{i+1}) = 0, \quad (1)$$
where $a_4 = a_1$.
Next we prove: if $x, y$ are distinct positive real numbers, and
$$m = \frac{f(x) - f(y)}{x - y}, \quad l = \frac{x f(y) - y f(x)}{x - y},$$
then for any $z$ satisfying $|x - y| < z < x + y$, we have
$$f(z) = m z + l \quad (2)$$
Indeed, from $|x - y| < z < x + y$ we know: there exist positive real numbers $a_1, a_2, a_3$ satisfying $a_1 + a_2 = x, a_2 + a_3 = y, a_3 + a_1 = z$. Substituting into equation (1) we obtain
$$\begin{aligned} 0 &= \sum_{i=1}^{3} (a_i - a_{i+1}) f(a_i + a_{i+1}) \\ &= (a_1 - a_2)(m x + l) + (a_2 - a_3)(m y + l) + (a_3 - a_1) f(z) \\ &= (a_1 - a_2)(m(a_1 + a_2) + l) + (a_2 - a_3)(m(a_2 + a_3) + l) + (a_3 - a_1) f(z) \\ &= m(a_1^2 - a_3^2) + l(a_1 - a_3) + (a_3 - a_1) f(z) \\ &= (a_1 - a_3)(m(a_1 + a_3) + l - f(z)) \end{aligned}$$
Also, since $x \neq y$, $a_1 - a_3 \neq 0$, therefore $f(z) = m(a_1 + a_3) + l = m z + l$, as claimed.

From equation (2) we immediately obtain: if $x, y$ are distinct positive real numbers, then when $t > |x - y|$,
$(t, f(t))$ lies on the line through the two points $(x, f(x)), (y, f(y))$. (3)
If $t \le |x - y|$, without loss of generality assume $x > y$. Then by (3), $(x, f(x))$ lies on the line through the two points $(y, f(y)), (t, f(t))$, so $(t, f(t))$ still lies on the line through the two points $(x, f(x)), (y, f(y))$.
From this we obtain that $f$ is a linear function, that is, there exist $a, b$ such that $f(t) = a t + b \forall t \in R^+$. But $f$ is a function from positive reals to positive reals, so $a, b > 0$.
Substituting back into the original equation, it is easy to verify that $f(t) = a t + b, (a, b > 0)$ is a solution of the original equation, therefore $f(t) = a t + b, (a, b > 0)$ are all the solutions of the equation.