Thanks to the feedback of @guangcun I have changed parts of this answer.
As I understand it now, the Drude weight distinguishes between an insulator and a metal in a clean model system. This means, for example, a lattice model with no impurities. The response is also calculated at $T=0$. These two things mean there are no dissipation mechanisms and so the currents do not decay, which is distinct from superconducting currents that are dissipationless at finite temperatures (below the critical temperature) and when there are some impurities.
How a delta in the frequency response relates to dissipationless currents:
Consider a simple 1D example without any consideration of spatial dependence of the response. An electric current $I(t)$ flows in response to a time-varying electric field $E(t)$. In linear response this current can be computed as
$$
I(t) = \int_{-\infty}^t dt' \, \sigma(t-t') \, E(t')
$$
where $\sigma(t-t')$ is the conductivity, and is also the Green's function of the current. Physically the Green's function of current is the current that will flow in response an impulse, $E(t) = E_0 \delta(t)$.
Now consider the what happens if $\sigma(\omega)$, the Fourier transform of $\sigma(t)$, contains a delta-function. Taking the inverse Fourier transform this gives a constant contribution to $\sigma(t)$
$$
\sigma(t) = \mathcal{F}^{-1}[ \, 2 \pi D \delta(\omega) \, + \, \ldots \,] = D \, + \, \ldots \, .
$$
Meaning, if the Drude weight is non-zero, an impulse produces a current that flows at all times without decaying. Or, considering a step function electric field, $E(t) = E_0 \theta(t)$ the current that flows
$$
I(t) = \int_{-\infty}^t dt' \, (D + \ldots) \, E_0 \, \theta(t') = D \, E_0 \, t + \ldots
$$
has a contribution that grows linearly with time.
