Charging circuit for automatically 'switching' between 1A and 2.4A?
iPhones and most other cell phones are charged off of a 5v USB cable connected to the phone, which also provides wired access to the phone to transfer data to ad from a PC.  Internally, the charging is controlled by a "battery management" chip like the bq24014.  This is certainly not the same chip as used in the iPhone; it just happens to be one I've used when designing a cellphone type device. This particular chip is only good for 1A below.

The battery management chip provides several functions.  First of all, it controls the charging through various stages -- first a constant current, then saturation phase with constant voltage until the charging current falls to a minimum level, and then turns the charging off.  During all of this, it monitors the battery temperature using an optional third lead coming from the battery.

With the bq24014, the maximum charging current is set using a resistor connect to the ISET pin.  Other battery management chips may use other schemes but the end result is the same.  Using a MOSFET and another resistor, it is possible to have more than one charging current possible,

Batteries are typically charged at 1C -- the mA capacity of the battery for one hour; but may be charged as low as .1C or as high as 2C or 3C (fast charge).

So as you can see, the charging current is totally controlled by the chip inside the device itself, and not by the power source (though of course the charging current can't be higher than what can be provided by the power source).   So if you step up the 3.6v from the 18650 using a boost regulator, and the charging current inside the phone is set at 2.4A, you're going to draw approximately 3.6A from the 18650:

$$2.4A \times \frac{(\frac{5v}{3.7v})}{0.9} = 3.6A$$

where 0.9 is the assumed efficiency of the boost regulator.

3.6A is just within the limits of the 2600 mAh 18650 battery, and well within the limits of a 3200 mAh battery, assuming a maximum discharge rate of 1.5C (yielding 3.9A and 4.8A respectively). Note as the voltage of the 18650 drops, the current draw will rise.  When the voltage of the smaller battery has dropped below 3.45v, the current will be near its maximum discharge level of 3.9A. Not too good a margin. But with the larger battery, this won't happen until the voltage drops below 2.8v.  Therefore: you need to use a 3200 mAh 18650 battery.