= Solution
Let $\mathcal G=-i[H,\cdot]+\sum_d\mathcal D[V_d]$ denote the <Lindbladian>. Substituting $\rho=\sum_n r_n\sigma_n$ and projecting with the <Hilbert-Schmidt inner product> gives
$$
\dot r_m=\sum_n\operatorname{Tr}[\sigma_m\mathcal G(\sigma_n)]r_n.
$$
For the <Quantum Hamiltonian> term, cyclicity of the <trace> yields
$$
\begin{aligned}
\operatorname{Tr}(\sigma_m[-iH,\sigma_n])
&=-i\operatorname{Tr}(\sigma_mH\sigma_n-\sigma_m\sigma_nH)\\
&=i\operatorname{Tr}[H(\sigma_m\sigma_n-\sigma_n\sigma_m)]
=L_{mn}.
\end{aligned}
$$
For each <Lindblad dissipator>, the same cyclic rearrangement gives
$$
\begin{aligned}
\operatorname{Tr}[\sigma_m\mathcal D[V_d](\sigma_n)]
&=\operatorname{Tr}(V_d^\dagger\sigma_mV_d\sigma_n)\\
&\quad-\frac12\operatorname{Tr}[V_d^\dagger V_d(\sigma_n\sigma_m+\sigma_m\sigma_n)]
=D_{mn}^{(d)}.
\end{aligned}
$$
These are the required coefficient matrices, so $\dot r=(L+\sum_dD^{(d)})r$. They are real because both generators map <Hermitian matrices> to <Hermitian matrices>, whose coordinates are real by (a).
The <trace> of a <commutator> vanishes. Also
$$
\operatorname{Tr}[\mathcal D[V](\rho)]
=\operatorname{Tr}(V^\dagger V\rho)
-\tfrac12\operatorname{Tr}(V^\dagger V\rho+\rho V^\dagger V)=0.
$$
Thus the last row of $L+\sum_dD^{(d)}$ is zero and $\boxed{\dot r_{N^2}=0}$. It is <trace> preservation, not preservation of the <purity of a density operator>, that produces this constant coordinate.
Write $G=L+\sum_dD^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt N$, the remaining coordinates obey the <Affine Bloch equation>
$$
\boxed{\dot s=As+c,\qquad A_{mn}=G_{mn},\quad
c_m=G_{m,N^2}/\sqrt N\quad(1\leq m,n\leq n).}
$$
Because $[H,I]=0$ and $\mathcal D[V](I)=[V,V^\dagger]$, its offset can also be written
$$
c_m=\frac1N\operatorname{Tr}\left[\sigma_m\sum_d[V_d,V_d^\dagger]\right].
$$
It vanishes for unital dynamics, but need not vanish for a general dissipative evolution.
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