= Solution
Write $q=|r-r'|$. Away from the source, the radial <Laplacian> gives $\nabla^2(-1/(4\pi q))=q^{-2}\partial_q[q^2/(4\pi q^2)]=0$. Through a small source-centred sphere, the outward flux of its gradient is
$$
\int_{q=\varepsilon}\partial_qH\,dS=\frac{1}{4\pi\varepsilon^2}4\pi\varepsilon^2=1.
$$
This identifies a unit <Dirac delta> at the source. More explicitly, for a smooth compactly supported test function $\psi$, apply <Green's second identity> outside the radius-$\varepsilon$ ball. The inner outward normal of that punctured domain is $-e_q$. The term $H\partial_n\psi$ has integral $O(\varepsilon)$, while $-\psi\partial_nH$ tends to $\psi(r')$. Thus $\int H\nabla^2\psi\,dV=\psi(r')$, proving the distributional identity
$$
\boxed{\nabla^2H(r,r')=\delta^{(3)}(r-r').}
$$
Whenever the integral converges, convolution with this <Green function of the Laplacian> solves the <Poisson equation for Newtonian gravity>:
$$
\boxed{\Phi(r)=-G\int_{\mathbb R^3}\frac{\rho(r')}{|r-r'|}\,d^3r'+h(r),\qquad\nabla^2h=0.}
$$
The harmonic term represents the homogeneous freedom and is fixed by boundary conditions. For an isolated finite mass distribution, regularity away from the prescribed sources and the normalization $\Phi\to0$ at infinity select $h=0$.
For a point mass at the origin, the delta integral gives the <Newtonian potential of a point mass>,
$$
\boxed{\Phi(r)=-\frac{GM}{|r|},\qquad r\ne0,\qquad\Phi(\infty)=0.}
$$
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