Why would an electronic timer have a minimum load?
The timer is designed to replace a regular switch as shown on the left of the schematic.







simulate this circuit &ndash; Schematic created using CircuitLab

Since there is often no neutral connection in the switch wall box it presents a problem for the timer designer. It won't work with a live feed only.

The trick is to replace the switch with a triac which behaves as a fast acting electronic switch and can switch on the current to the lamp at any point in the half-cycle.



Triac modified sinewave. Delaying the turn-on point results in reduced voltage on the lamp.

If the timer circuit delays the turn-on point of the triac the voltage to the lamp is reduced. That leaves the remainder available to power the switch. The minimum 40 W lamp ensures that there is enough of a load to pass enough current to keep the timer running. The designers hope you won't notice that the lamp is ever so slightly dimmer.

When the lamp is off there is almost full voltage across the triac so the timer again has power.

These circuits are a little tricky to design as they have to work with full voltage and a fraction of supply voltage.

Note that if the bulb is removed that there is no return path for the timer so it will power down - although there may be a delay if it has some energy stored in a capacitor.

Battery

The battery is probably to keep the clock running during power cuts. As explained in the answer it can always steal enough power when the supply is there. The trigger delay is exaggerated in my sketch and probably 10 V or so would be enough to charge up a capacitor in the switch. The low voltage portions of the sine wave contribute very little to the power of the lamp. On a 240 V supply \$V_peak_ = \sqrt 2 \times 240 = 340 V\$. Since power is proportional to the square of the voltage, the power to the lamp is \$\frac{340^2}{10^2} = 1150 \$ times stronger at peak than at 10 V.