= Solution
For a degree-$n$ <homogeneous polynomial>, the <Euler homogeneous function theorem> gives $\mathbf r\cdot\nabla H_n=nH_n$. On $r>0$ in three-dimensional <Euclidean space>,
$$
\nabla r^s=s r^{s-2}\mathbf r,\qquad \Delta r^s=s(s+1)r^{s-2}.
$$
Apply the <product rule for the Laplacian>:
$$
\Delta(r^sH_n)
=r^s\Delta H_n+2s r^{s-2}\mathbf r\cdot\nabla H_n+s(s+1)r^{s-2}H_n
=r^s\Delta H_n+s(2n+s+1)r^{s-2}H_n.
$$
For $s=-(2n+1)$ the second term vanishes. Hence <homogeneous harmonic inversion> gives
$$
\boxed{\Delta\!\left(\frac{H_n}{r^{2n+1}}\right)
=\frac{\Delta H_n}{r^{2n+1}},\qquad r>0.}
$$
The nonzero factor $r^{-(2n+1)}$ proves both directions: the transformed function is <harmonic> exactly when $\Delta H_n=0$ away from the origin. Since $\Delta H_n$ is a polynomial, vanishing on the punctured space makes it identically zero everywhere.
Equivalently, homogeneity shows that the <Kelvin transform> is
$$
K_aH_n(\mathbf r)=\frac arH_n\!\left(\frac{a^2}{r^2}\mathbf r\right)
=a^{2n+1}\frac{H_n(\mathbf r)}{r^{2n+1}}.
$$
It sends a <regular solid harmonic> to an <irregular solid harmonic>, exchanging the radial powers $r^n$ and $r^{-n-1}$. The origin is excluded from the harmonicity statement: for example, the $n=0$ transform $1/r$ is harmonic only away from zero and has a distributional <Dirac delta function> source there.
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