Since $f$ is Riemann integrable on any finite interval, it is bounded on any finite interval. It then follows from (ii) applied to $n = 1$ that $f$ is continuous. The set of functions satisfying the two properties form a subvector space of the space of all continuous functions. It is clear that all linear functions belong to this subspace. We will show that there is no other function. Let $f$ satisfy the two properties. By adding a linear function to $f$, we may assume that $f(0) = f(1) = 0$. It suffices to show that $f = 0$. We first prove a Lemma: If $f(a) = f(b) = 0$ for some $a < b$, then we have $f(x) = 0$ for all $x \in [a, b]$. Proof: Suppose this is not the case. Since $f$ is continuous, we may assume without loss of generality that $m= \max\{f(x):x \in[a, b]\}$ is strictly positive. We consider the set $C = \{x \in [a, b]: f(x) = m\}$. This is a nonempty closed subset of $\Bbb R$. However, for any $c \in C$, we must have $c \in (a, b)$ and hence there exists $n$ such that $[c - \frac 1n, c+\frac1n] \subseteq[a, b]$. We have $m = f(c) = \frac n 2 \int_{c - \frac 1 n}^{c + \frac 1 n} f(t)dt$ and also $m \geq f(t)$ for all $t\in [c - \frac 1n, c+\frac1n]$. These imply that $f(x) = m$ for all $x \in (c - \frac 1 n, c + \frac 1 n)$. We have shown that, for any $c \in C$, there exists an open interval $I$ such that $c \in I \subseteq C$. This means that $C$ is an open subset of $\Bbb R$. Therefore $C$ is both open and closed, which is impossible, as $\Bbb R$ is connected. Since $f(0) = f(1)$, the Lemma implies that $f(x) = 0$ for all $x \in [0, 1]$. This means that the set $B = \{z: f(x) = 0, \forall 0 \leq x \leq z\}$ is nonempty. We let $b$ denote the supremum of the set $B$. We want to show that $b = \infty$. Suppose it is not the case. Then we have $f(x) = 0$ for all $0 \leq x \leq b$. By condition (ii) applied with $n = 1$, we have $0 = f(b) = \frac12\int_{b - 1}^{b + 1}f(t)dt = \frac12\int_b^{b + 1}f(t)dt$. Since $f$ is continuous, this implies that $f(c) = 0$ for some $c \in (b, b+1)$. Then the Lemma tells us that $f(x) = 0$ for all $x \in [b, c]$. This leads to $c \in B$, contradicting the assumption that $b = \sup B$. Therefore we have $\sup B = \infty$, i.e. $f(x) = 0$ for all $x \geq 0$. The same argument, by considering $A = \{z:f(x) = 0, \forall z \leq x \leq 1\}$ and $a = \inf A$, shows that $f(x) = 0$ for all $x \leq 1$. Thus we have proved $f = 0$. In conclusion, all functions satisfying the two properties are the linear functions.

The answer is all functions of the form $f(x)=cx+d$ where $c,d$ are arbitrary real numbers. Just as above, one can show that $f$ is continuous. Using the Fundamental Theorem of Calculus we have that \[ f'(x)= \frac{n}{2}\left(f\left(x+\frac{1}{n}\right)-f\left(x-\frac{1}{n}\right)\right)\qquad \qquad (\text{1}) \]which implies that $f$ is infinitely differentiable at each point. Taylor expanding $(\text{1})$ we find that \begin{align*} f'(x)&=\frac{n}{2}\left(f(x)+f'(x)\frac{1}{n}+f''(x)\frac{1}{4n^2}+f^{(3)}(x)\frac{1}{6n^3}+o\left(\frac{1}{n^3}\right)-f(x)+f'(x)\frac{1}{n}-f''(x)\frac{1}{4n^2}+f^{(3)}(x)\frac{1}{6n^3}+o\left(\frac{1}{n^3}\right)\right)\\ &=f'(x)+f^{(3)}(x)\frac{1}{6n^2}+o\left(\frac{1}{n^2}\right) \end{align*}hence $f^{(3)}(x)=o(1)$, which implies that $f^{(3)}(x)=0$ and therefore $f(x)$ must be a quadratic polynomial. Checking the initial condition it is easy to see that $f$ must be linear.