If $u:\mathbb{C}\to \mathbb{R}$ is continuous and satisfies $u(z)=\frac{1}{2\pi}\int_0 ^{2\pi}u(z+\frac{e^{i\theta}}{n})d\theta$ for all $n\in \mathbb{N}$ and $z\in \mathbb{C}$, is $u$ harmonic?
What I know: If for each $P\in \mathbb{C}$ there exists $r_P$ such that for $0<r<r_P$, $u(P)=\frac{1}{2\pi}\int_0 ^{2\pi } u(P+re^{i\theta})d\theta$ then $u$ is harmonic, but in the proof I need the fact that the equation holds for all $0<r<r_P$.
I tried showing that $u$ is a limit of holomorphic functions satisfying the integral equation for $0<r<r_P$, and I considered looking for a counterexample but I don't know where to start.
Yes, a continuous function with the "discrete mean value property" is harmonic.
Such a function satisfies the maximum (and minimum) principle in the form
Postponing the proof of this, we use it to deduce the harmonicity:
Fix an arbitrary $z_0\in \mathbb{C}$, and let $h$ be the solution to the Dirichlet problem with boundary values $u$ on $D_1(z_0)$, that is, $h\colon \overline{D_1(z_0)} \to \mathbb{R}$ is continuous, harmonic on the open disk $D_1(z_0)$ and satisfies $h(z_0 + e^{it}) = u(z_0+e^{it})$ for $0\leqslant t \leqslant 2\pi$. Then $v = u - h$ is a continuous function on $\overline{D_1(z_0)}$ that satisfies the maximum (and minimum) principle in the above form, and vanishes on the boundary. If $v$ were not identically $0$, it would attain a global maximum or minimum $M \neq 0$ in the interior of the disk, and the set $\{ z\in D_1(z_0) : v(z) = M\}$ would be open by the maximum/minimum principle, and closed by the continuity of $v$, hence $v\equiv M$, but that contradicts the vanishing boundary values. Thus $u \equiv h$ is harmonic on $D_1(z_0)$. Since $z_0$ was arbitrary, $u$ is harmonic everywhere.
Note that the function must satisfy both, the maximum and the minimum principle in the above form, if only one were satisfied, subharmonic resp. superharmonic functions would give counterexamples.
Now to the proof that functions with the DMVP satisfy the above maximum principle:
Suppose $u$ has a local maximum in $z_0$, and $r > 0$ is such that
$$u(z_0) = \max \{ u(z) : \lvert z-z_0\rvert \leqslant r\}.$$
For all $n > \frac{1}{r}$, $d_n \colon t \mapsto u(z_0) - u\bigl(z_0 + \frac{1}{n}e^{it}\bigr)$ is a non-negative continuous function, and by the DMVP,
$$\int_0^{2\pi} d_n(t)\,dt = 0,$$
hence $d_n \equiv 0$. Let $n = \lceil 1/r\rceil + 1$. For each $z$ with $\lvert z-z_0\rvert = \frac{1}{3n}$, we have
so it follows that $u(w) = u(z) = u(z_0)$ for all $w$ with $\lvert w-z\rvert = \frac{1}{2n}$. But that means we have $u(w) = u(z_0)$ for all $w$ in
$$\bigcup_{\lvert z-z_0\rvert = \frac{1}{3n}} \left\{ w : \lvert w-z\rvert = \frac{1}{2n}\right\} = \overline{D_{5/(6n)}(z_0)},$$
showing that the maximum principle holds. The minimum principle is proved completely analogously, or by observing that $u$ has the DMVP if and only if $-u$ has it.
It is not relevant that the radii for which the mean value property is assumed are the reciprocals of the natural numbers, the same argument works for all sequences of radii $r_n \searrow 0$.