Continuing on from the numerical analysis thread...

Can solve for [$]I_i[$]:

[$]I_i(t)=e^{-\lambda t}\left( \lambda \int_0^t e^{\lambda \tau}S_i(\tau)d\tau-e^{\lambda t}S_i(t)+S_i(0)+I_i(0) \right).[$]

And then

[$]\frac{dS_i}{dt}=\left(\alpha-\frac{\beta}{N} \right)S_i\left( S_i- \lambda \int_0^t e^{\lambda (\tau-t)}S_i(\tau)d\tau -e^{-\lambda t}(S_i(0)+I_i(0))\right)\\ \qquad + \frac{\beta}{N}S_i\left( S- \lambda \int_0^t e^{\lambda (\tau-t)}S(\tau)d\tau -e^{-\lambda t}(S(0)+I(0))\right). [$]

Observe where there are and aren't subscripts.

When [$]\beta=0[$] there is total lockdown and you can integrate each [$]S_i[$] a bit. (*) Then you need to sum over all [$]i[$] and you'll get [$]I[$] which feeds into the [$]H[$] and [$]D[$] equations. Messy. So that's solve first then sum over [$]i[$].

When [$]\alpha=\frac{\beta}{N}[$] there is no lockdown. You can sum the [$]S_i[$] equations over all [$]i[$] and get the same result as in (*) but for the sum instead of individual households. So that's sum over [$]i[$] then solve. (The other way around from above.)

For other cases you can't solve/sum in either order because of the [$]S_i^2[$] term.