Solution:
We may assume that $AC < BC$. Since $\text{ACI}$ and $\text{AMI}$ are acute, then $\triangle ACI \cong \triangle AMI$.
Hence $AC = AM$ and $\text{AIC = AIM}$, i.e.
$$\text{CIM = 360 - 2 AIC = 180 - ABC.}$$
Note that $\text{ABC}$ is maximal if $BC$ is tangent to the circle with center $A$ and radius $AM$.

Then $\text{ACB = 90}$, $\text{ABC = 30}$, and hence the least possible value of $\text{CIM}$ is $150^\circ$.