I can understand the confusion. The high voltage cells are not being discharged. They are just being charged at a slower rate than the others to minimise the voltage delta (as T1 explained above) - see my diagram below for a (hopefully) clearer explanation.
almost... but not quite
No... that is impossible in a series circuit that is being charged.
As @Coulomb explained, the BMS will switch a 'bleed' resistor in parallel with one or more 'higher-voltage' cells during charging.
Thanks to Kirchoff's current law, the total current entering a node must equal the total current leaving it. So in the diagram below
Icharge =
Icell +
Ibleed.
When a cell reaches a higher voltage than the others, the BMS activates a solid state switch that connects a shunt/bleed resistor in parallel with that specific cell.
Because a portion of the current is diverted through the bleed resistor the current entering the 'highest-voltage' cell is reduced, allowing other lower-charged cells to catch up hopefully before the high-voltage cell reaches its peak as some current ( the I-cell element) will still be charging it.
View attachment 44697