= Solution
First suppose $u$ is a classical <harmonic function>. At a local maximum of $u-v$, its <Hessian matrix> is <negative semidefinite>, so $0=\Delta u(x_0)\leq\Delta v(x_0)$. At a local minimum the Hessian inequality reverses and gives $\Delta v(x_0)\leq0$. Thus both test conditions hold.
Conversely, suppose only that $u$ is continuous and satisfies the two conditions. We prove the stronger conclusion that these conditions themselves imply $u\in C^\infty$ and $\Delta u=0$. Use the standard Dirichlet existence theorem on a ball: every continuous function on its boundary has a harmonic extension $h\in C^\infty(B_R(a))\cap C(\overline{B_R(a)})$ with exactly those boundary values. This follows, for example, from the Poisson integral. Apply it to $u$ on an arbitrary <relatively compact> ball, obtaining a <harmonic replacement> $h$.
If $u-h$ is positive somewhere, choose $\varepsilon>0$ small enough that $u-h-\varepsilon(R^2-|x-a|^2)$ is still positive there. It vanishes on the boundary and therefore has a positive maximum at an interior point. The upper test
$$
v=h+\varepsilon(R^2-|x-a|^2)
$$
has $\Delta v=-2n\varepsilon<0$, contradicting the maximum test condition. Similarly, if $u-h$ is negative somewhere, $v=h-\varepsilon(R^2-|x-a|^2)$ yields an interior negative minimum of $u-v$ for small $\varepsilon$, but $\Delta v=2n\varepsilon>0$, contradicting the minimum condition. Hence $u=h$ throughout the ball.
If the quantifier requires $v\in C^2(\Omega)$ rather than a local test, multiply the displayed smooth $v$ by a smooth cutoff supported within the ball and equal to one near the touching point, extending by zero outside. The resulting global test has the same germ and <Laplacian> there, so the argument remains valid. Since every point lies in such a ball, $\boxed{u\text{ is a smooth harmonic function throughout }\Omega}$. This is the <continuous viscosity harmonic functions are classical> characterization.
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