= Solution
For $F\dashv G$ with unit $\eta$ and counit $\varepsilon$, the <monad induced by an adjunction> is
$$
T=GF,\qquad \mu=G\varepsilon F,
$$
with unit $\eta$. An <algebra for a monad> is $(A,a:TA\to A)$ satisfying $a\eta_A=1_A$ and $aT(a)=a\mu_A$.
The <Eilenberg-Moore comparison functor> is
$$
K:\mathcal D\to\mathcal C^T,\qquad K(D)=(GD,G\varepsilon_D).
$$
If $\mathcal D$ has coequalizers of reflexive pairs, define
$$
FTA\mathrel{\substack{\xrightarrow{Fa}\\[-2pt]\xrightarrow[\varepsilon_{FA}]{} }}FA
\xrightarrow{q}L(A,a)
$$
for a $T$-algebra $(A,a)$. The pair has common section $F\eta_A$. Maps $L(A,a)\to D$ correspond by the coequalizer property and the adjunction exactly to algebra morphisms $(A,a)\to K(D)$, naturally in both variables. Hence this is the <Left adjoint to the Eilenberg-Moore comparison functor>. The <monadic length> is the least number of successive comparison steps required for the resulting monadic tower to become an equivalence.
For $A\in\mathcal C_n$, let $A_n=\{a:\alpha_n(a)=a\}$. The free $\mathcal C_{n+1}$-object has underlying set
$$
B=A\times\{0\}\;\cup\;A_n\times\mathbb N_{>0}.
$$
Retain all old operations on $A\times\{0\}$. Put
$$
\alpha_{n+1}(a,0)=(a,1)\quad(a\in A_n),
$$
and on every new chain put
$$
\alpha_1(a,k)=(a,k+1).
$$
For $i>1$, $\alpha_i$ is undefined on the new points because none is fixed by $\alpha_1$. These definitions satisfy the domain conditions. If $f:A\to U(B')$ is a $\mathcal C_n$-morphism, its unique extension sends
$$
(a,k)\longmapsto\beta_1^{\,k-1}\bigl(\beta_{n+1}(f(a))\bigr)
\quad(k>0).
$$
This proves the required left adjoint.
After adjoining $\alpha_{m+1}$ freely, the newly added points lie on fixed-point-free $\alpha_1$-chains, while each old point where $\alpha_{m+1}$ was added is no longer fixed by $\alpha_{m+1}$. Consequently there are no points at which a further free operation must be adjoined. Thus for every $n>m$ the endofunctor and unit/multiplication of the monad induced on $\mathcal C_m$ are already those induced by $\mathcal C_m\rightleftarrows\mathcal C_{m+1}$.
By assumption each adjacent adjunction is a <monadic adjunction>. The first comparison for $\mathcal C_m\rightleftarrows\mathcal C_n$ therefore recovers $\mathcal C_{m+1}$, and iteration successively recovers $\mathcal C_{m+2},\ldots,\mathcal C_n$. None of the intervening forgetful functors is an equivalence, since the next partial operation can be chosen differently on a fixed point. Starting at $\mathcal C_0$ takes exactly $n$ steps:
$$
\boxed{\text{the monadic length of }\mathcal C_0\rightleftarrows\mathcal C_n\text{ is }n.}
$$
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