Impedance of an LCR series circuit can be maximum when either Inductive Reactance or Capacitive Reactance is maximum at an instant of time.
Max Capacitive Reactance is at minimum frequency, and max Inductive Reactance is a max frequency. So, which is the condition for maximum Impedance?
The answer is very simple. The total complex impedance Z is $$Z = R + j𝛚L + 1/(j𝛚C)$$ where $R$ is the ohmic resistance, $j$ is the imaginary unit, $𝛚$ is the angular frequency, $L$ is the inductance, and $C$ is the capacitance. The second and third term on the RHS is the complex inductor and capacitor impedance, respectively.
The maximum impedance can only mean the maximum of the absolute value of the impedance Z: $$\lvert{Z}\rvert = \sqrt{ R^2 + (\omega L - 1/ \omega C)^2}$$ Thus the impedance $$\lvert{Z}\rvert→∞$$ as $\omega L$ for $\omega$ $→∞$. It increases indefinitely for increasing frequency. The impedance also goes to infinity for $\omega$ $→0$. Therefore, there is no relative maximum of the impedance.
There is, however, a minimum impedance $\lvert{Z}\rvert = R$ at the resonance frequency $\omega = 1/\sqrt{LC}$.
The "conditions" are circuit dependent. Sometimes there may not even be a maximum (such as a single capacitor, whose impedance goes to infinity as frequency goes to zero). The most common way to analyze this is to treat the circuit as a 2 port model, with an input voltage between two wires and an output voltage between two other wires. Once we think about the circuit this way, the maximum impedance is the frequency where the gain of this circuit is at a minimum.
You can do this by hand in the time domain, but usually we take advantage of the fact that the LCR circuit is linear and use the Laplace transform which converts the time domain description of this circuit (with all its derivatives and integrals) into an algebraic formula in terms of a "complex frequency" (usually denoted with $s$). This is very convenient because finding the minimum of an algebraic equation is much simpler than finding the minimum of a complicated mix of derivatives. One finds the $omega$ such that when one substitutes $[s:= j\omega]$ into that algebraic expression, it's a minimum. That frequency is the frequency where the minimum occurs, and the resulting gain can be used to compute the impedance at that point.
Alternatively, if you haven't learned Laplace transforms yet, it may be easiest to put it into a circuit analyzer and just have it plot the response at different frequencies. But if you want to do it by hand, Laplace transforms are the tool of choice to do it as easily as possible.
