= Solution
For an integrable applied voltage, the <convolution theorem> gives
$$
\widetilde I(\omega)=\left[\frac{a_1p_1}{(\gamma_1+i\omega)^2+p_1^2}
+\frac{a_2p_2}{(\gamma_2+i\omega)^2+p_2^2}\right]\widetilde V(\omega).
$$
The integrated <electric current> is the zero-frequency transform, $Q=\widetilde I(0)$. Since $\widetilde V(0)$ can be arbitrary, zero total <electric charge> for every such input is equivalent to
$$
\boxed{\frac{a_1p_1}{\gamma_1^2+p_1^2}+\frac{a_2p_2}{\gamma_2^2+p_2^2}=0.}
$$
For nonzero $p_1,p_2$, this can be written
$$
\boxed{\frac{a_1}{a_2}=-\frac{p_2(\gamma_1^2+p_1^2)}{p_1(\gamma_2^2+p_2^2)}.}
$$
Because both amplitudes are positive, nonzero $p_1,p_2$ must have opposite signs. If exactly one $p_i$ is zero there is no cancellation with positive amplitudes; if both vanish the kernel is identically zero and any amplitudes work. In particular, if both frequency parameters were additionally required to be positive, no such nontrivial system would exist. The PDF requires positivity of $a_i,\gamma_i$, but allows real fixed $p_i$.
This is <zero-frequency cancellation of a causal response kernel>. The phrase arbitrary voltage is understood within a class for which the net-charge integrals exist; a permanent DC input need not have finite total charge.
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