Finite sectional event dynamics preserve positivity and retained-plus-reservoir mass

At every nonnegative state with $n_i=0$, no event consuming coordinate $i$ has positive rate, because each such rate contains $n_i$; all events producing $i$ have nonnegative rates and yields. Thus $F_i(n,L)\ge 0$ on that face of the orthant.

For the ODE, extend the event formulas to all real states by the same finite polynomials; they agree with the algorithm on admissible states and are locally Lipschitz, since differences of quadratic monomials are bounded by a constant times the coordinate differences on each ball. On a compact time interval in the solution's existence interval choose a Lipschitz constant $C$ on a ball containing both the solution and its coordinatewise positive part. Set $n_i^{+}=\max (n_i,0)$ and $u(t)={\sum }_i\max (-n_i(t),0)$. For each coordinate with $n_i<0$, one has $n_i^{+}=0$, so step 1.1 and Lipschitz continuity give $F_i(n,L)\ge F_i(n^{+},L)-C\mathrm{\parallel }n-n^{+}{\mathrm{\parallel }}_2\ge -Cu$. The absolutely continuous negative-part chain rule gives $(\max (-n_i,0){)}^{\mathrm{\prime }}=-F_i$ when $n_i<0$, zero when $n_i>0$, and zero almost everywhere on the level set $n_i=0$. Hence $u^{\mathrm{\prime }}(t)\le (M+1)Cu(t)$ almost everywhere. Since $u(0)=0$ and $u\ge 0$, differentiating $e^{-(M+1)Ct}u(t)$ shows it is nonincreasing from zero, so $u(t)=0$. Once $n\ge 0$, the reservoir derivative is a sum of nonnegative overflow rates, hence $L(t)\ge 0$.

For an in-range aggregation event with parent masses $x_i,x_j$, the pivot output has mass $x_i+x_j$ by [F2]; subtracting the two parents gives zero augmented-mass change. For overflow, the retained parents remove mass $x_i+x_j$ and the algorithm adds exactly this amount to $L$. For fragmentation of parent $x_i$, the daughter measure has first moment $x_i$ by the algorithm input, and each pivot projection preserves daughter mass by [F2], so parent removal and daughter addition cancel. Thus every event increment has zero augmented-mass change. Pairing the ODE with the mass vector and adding $\dot{L}$ gives ${\dot{m}}_{\mathrm{t}\mathrm{o}\mathrm{t}}=0$; integrating yields $m_{\mathrm{t}\mathrm{o}\mathrm{t}}(t)=m_{\mathrm{t}\mathrm{o}\mathrm{t}}(0)$.