Every time ADC samples, it draws charge from your input capacitor. I think this is the basic thing you missed. The ADC draws a current from the input signal during sampling, and if the input cannot supply that current, the voltage sags and you get a wrong result. OTOH, if you have a ceramic capacitor here, it can easily supply the current required, but when the ADC discharges the capacitor, its voltage level changes.
Sure, if the input capacitor is very large, say 10000x larger than the ADCs sampling capacitor, the change in your input capacitor voltage level is only ~0.01%.
But, if you take multiple samples, the input capacitor loses a tiny bit of charge every time the ADC samples. So this capacitor needs to be topped off with the fresh input voltage. The 2.2kOhm resistor limits this current.
It's likely that your 2.2kOhm was too much for the sampling rate you use. Using smaller resistor charges the input capacitor faster. Another option is to sample less frequently, so that the capacitor has time to replenish to the correct input value through that 2.2kOhm.
Note there are two different time constraints here:
Assuming Cin >> Csampl, the low-impedance requirement for the ADC is satisfied by the capacitor: the very short sampling time becomes irrelevant, the capacitor handles it. Then, the resistor must be sized so that the period between samples (1/samplerate) is catered by that resistor.
If Cin >> Csampl isn't true - for example, no Cin at all, or Cin which is only some tens times of larger than Csampl - the input resistance must be even smaller, because now it needs to be able to supply the charge ADC requires during the sampling time, which can be something like a microsecond.
For example:
If you sample every 1ms, and sampling takes 1us, and each sampling extracts 10pC of charge from the input signal:
If you have no input capacitor (or small input capacitor), the input must be able to supply 10pC during 1us (during sampling)
If you have a large input capacitor (say 100nF), it supplies 10pC to the ADC during that 1us. The input signal now needs to supply 10pC to that capacitor during 1ms.