Bay Area Mathematical Olympiad

Solution:
We will use the notation $f^{n}(x)$ to denote $f(f(\cdots f(x) \cdots))$, where we iterate the function $n$ times. Suppose, to the contrary, that $f(x) \neq x$ for some $x \in S$. This implies that $f(f(x)) \neq f(x)$ as well, since $f(f(x))$ is either the sum or the average of $x$ and $f(x)$ and these are distinct non-zero real numbers. Likewise, $f(f(x)) \neq f(x)$ implies that $f^{3}(x) \neq f(f(x))$. We can keep iterating the function to create a sequence
$$x, f(x), f(f(x)), f^{3}(x), f^{4}(x) \cdots$$
where no term is equal to the term preceding it. However, since $S$ is finite, eventually there has to be a repeating value. In other words, there exists $m, n$, with $n>m+1$, such that $f^{m}(x)=f^{n}(x)$.
Let $a=f^{m}(x)$. Then $a, f(a), f(f(a)), \cdots, f^{n-m}(a)=a$ is a cycle of length $n-m$. Since $f(f(x))$ cannot equal $x$ (the sum of $x$ and $f(x)$ cannot equal $x$ since $f(x)$ is nonzero and the average of $x$ and $f(x)$ cannot equal $x$ because $f(x) \neq x$ ), the cycle has length at least 3. Since the cycle is finite, and the terms are nonzero, there must be a term of maximum absolute value. Call this $M$, and without loss of generality, assume that $M$ is positive.
We know that the cycle has at least three terms, so consider the consecutive terms $U, V, M$ in the cycle (since it is a cycle, it can start "anywhere"). We have $V=f(U)$ and $M=f(V)=f(f(U))$. We claim that $V$ is positive, for if it were negative, then $M$ would be either the average of $U$ and $V$ or the sum of $U$ and $V$, which would force $U$ to be larger than $M$, contradicting the fact that $M$ is the largest term in the cycle.
But if $V$ is positive, then $f(M)=f(f(V))$ must be greater than $M / 2$, since it is either the sum or average of a positive number and $M$. Likewise, $f(f(M))$ must also be greater than $M / 2$, since it is either the sum or average of $M$ and a value that is greater than $M / 2$. Once we have two consecutive terms in the cycle that are greater than $M / 2$, all subsequent terms in the cycle will be greater than $M / 2$. In other words, the cycle starting at $M$,
$$M, f(M), f(f(M)), f^{3}(M), \cdots$$
consists entirely of terms whose value is greater than $M / 2$. Also, starting with the third term, each term is either the sum or average of the two terms preceding it. But since it is a cycle, eventually it will come back to the value of $M$, and that is impossible: $M$ is neither the sum nor the average of two terms greater than $M / 2$. We have achieved a contradiction, and conclude that there are no $x \in S$ such that $f(x) \neq x$; i.e. $f(x)=x$ for all $x \in S$.