= Solution
Besides the stated scales, set $P=S_0p$, $\epsilon=E_0/S_0$, $\alpha_i=k_i/k_1$, $\beta_i=(k_{-i}+k_i^+)/(k_1S_0)$, and $\kappa_i=k_i^+/(k_1S_0)$. Then $r_i=\alpha_ic_{i-1}s-\beta_ic_i$, and direct substitution, remembering $d/dt=k_1E_0d/d\tau$, gives
$$
\boxed{\epsilon c_{0,\tau}=-r_1,\quad\epsilon c_{i,\tau}=r_i-r_{i+1}\ (1\leq i<n),\quad\epsilon c_{n,\tau}=r_n,}
$$
$$
\boxed{s_\tau=-\sum_{i=1}^n(r_i+\kappa_ic_i),\qquad p_\tau=\sum_{i=1}^n\kappa_ic_i,\qquad\sum_{i=0}^nc_i=1.}
$$
The exact initial values are $s(0)=1$, $p(0)=0$, $c_0(0)=1$, $c_i(0)=0$ for $1\leq i\leq n$. The enzyme-state equations carry the small parameter because the <enzyme> <concentration> scale is small relative to the initial <substrate> <concentration>; the <substrate> evolves on the chosen slower depletion time scale.
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