A field effect transistor (FET) is a voltage sensitive device with a very high input impedance (unlike a bipolar junction transistor BJT). Hence it is very sensitive to signals of very low currents and voltages.
Junction field-effect transistor (JFETS) generally also have a lower intrinsic noise level than a MOSFET or BJT which, in addition to the high input impedance, makes them a good choice for amplifying very small signals.
Jfets are specified at zero volts, gate to source. The 4416 has an idle current of 5 to 15 milliamps at 0Vgs and that is how this circuits starts. The 4416 allows current into R2, the 1k resistor, and a positive voltage develops on the top of R2. This is going in the direction to stop current flow through the jfet. The cutoff voltage for a 4416 is between 2 and 6 volts with the source more positive than the gate. In this circuit, the gate is stuck at zero volts by the 47M resistor. As the current through the 1k resistor develops a voltage, the jfet circuit finds a balance and settles on that amount of current.
If the jfet circuit settles at 3 ma, there will be 3 volts across the 1k resistor. That voltage will hold the base of TR1 at 3 volts. Meanwhile, R6 and R7 are holding the other base of the differential pair at half the battery voltage...4.5 volts. TR2 allows current through R5 such that there is no current through TR1.
When the "sensing tip" is connected to a positive voltage the jfet allows more current. When that current exceeds 4.5 ma, TR1 comes on and the LED lights up.
A bit more than you asked, but it's an easy circuit.
R3 and R4 could be removed but, as they are now, they reduce the effects of temperature and batch differences on the 2 transistors. They don't seem important unless you are going to add an adjustment to set this thing to maximum sensitivity. Then you would need it stable and predictable, like the next guy says.