= Solution
The <Gauss map> of an oriented surface is its smooth unit normal $N:S\to S^2$. With the convention that the <shape operator> is $A_p=-dN_p$, the <second fundamental form> is $\mathrm{II}_p(v,w)=\langle A_pv,w\rangle$ and the <normal curvature> in a nonzero direction $w$ is $k_n(w)=\mathrm{II}(w,w)/\langle w,w\rangle$. The two <principal curvatures> are the real eigenvalues $k_1,k_2$ of this self-adjoint operator; the <mean curvature> is $H=(k_1+k_2)/2$.
For directions successively spaced around the tangent plane, write their angles to a principal direction as $\theta_j=\theta_0+(j-1)\pi/m$. The <Euler formula for normal curvature> gives
$$
\widetilde k_j=k_1\cos^2\theta_j+k_2\sin^2\theta_j=H+\frac{k_1-k_2}{2}\cos(2\theta_j).
$$
The <geometric series> $\sum_{j=0}^{m-1}e^{2\pi ij/m}=0$ for $m\geq2$ makes the cosine sum zero. Therefore
$$
\boxed{\sum_{j=1}^m\widetilde k_j=mH.}
$$
Here equal angular spacing is understood in the usual directed sense. If the printed consecutive angles were only unsigned, without requiring the same rotational sense, the assertion would be false: for $m=3$ the directions $0,\pi/3,0$ satisfy those unsigned angle conditions but do not generally give $3H$.
A <minimal surface> has $H=0$, so at each point $k_2=-k_1$. In an orthonormal principal basis, $A=\operatorname{diag}(k_1,-k_1)$ and $A^2=k_1^2I$. Hence
$$
\langle dN_p(w_1),dN_p(w_2)\rangle=\langle A_pw_1,A_pw_2\rangle=k_1^2\langle w_1,w_2\rangle,
$$
which proves the identity with $\boxed{\mu(p)=k_1^2=-K(p)\geq0}$, where $K$ is the <Gaussian curvature>.
The converse is false. The identity only forces $k_1^2=k_2^2$, which allows either $k_2=-k_1$ or $k_2=k_1$. A sphere of radius $R$ has $dN=I/R$ for its outward normal, so it satisfies the identity with $\mu=R^{-2}$, yet $H=-1/R\ne0$ under the chosen convention. Thus a conformal Gauss map can also arise from an umbilic nonminimal surface.
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