= Solution
Add a fourth level using
$$
\dot X_3=X_3(b_2X_2-d_3-a_3X_4),\qquad
\dot X_4=X_4(b_3X_3-d_4),
$$
with the first two equations unchanged. The positive <equilibrium> must be
$$
\boxed{X_{2*}=\frac r{a_1},\quad X_{3*}=\frac{d_4}{b_3},\quad
X_{1*}=\frac{d_2+a_2d_4/b_3}{b_1},\quad
X_{4*}=\frac{b_2r/a_1-d_3}{a_3}.}
$$
The first three entries are positive automatically. The fourth is positive exactly when $b_2r/a_1>d_3$. Thus the printed coexistence assertion requires adequate herbivore production to support the third and fourth levels; it is not true for every positive parameter set. The top predator supplies the extra mortality that was absent in the unbounded three-level case.
For a chain of $N$ levels, the same equations give $b_{i-1}X_{i-1}-d_i-a_iX_{i+1}=0$ in the interior, with the basal and top equations fixing alternating levels. The interaction <matrix> has zero diagonal and opposite-sign nearest-neighbor entries. Its <determinant> is nonzero for even $N$ and zero for odd $N$: expansion at the end gives $\det A_N=a_{N-1}b_{N-1}\det A_{N-2}$, with $\det A_0=1$, $\det A_1=0$. For even $N$ this gives a unique <equilibrium> candidate, but its entries still need to be positive. For odd $N$ a compatibility condition is necessary, just as for the three-level example. This is <Lotka-Volterra food-chain parity>. The parity result is specific to an exponential unregulated resource and nearest-neighbor trophic interactions.
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