Low-cost, low-current solution for dropping 1V from a battery
This microchip appnote suggests in section 8-3 to use some rectifier diodes in series as shown in Figure 3-1, which converts a 5V supply into an approximately 3.9V supply.



Each diode will give you a voltage drop, depending on the type of the diode and the current  through the diode.  Note the inclusion of R1, which the appnote mentions 


  is present to keep the voltage at the PIC MCUs \$V_{DD}\$ pin 
  from exceeding the maximum \$V_{DD}\$
  at minimum loads (typically when the PIC MCU 
  is in Reset or sleeping). Depending on the 
  other circuitry connected to \$V_{DD}\$, this resistor 
  may have its value increased or possibly even 
  eliminated entirely. Diodes D1-D3 must be 
  selected so that at maximum load, typically
  when the PIC is running and is driving its 
  outputs high, the voltage drop across D1-D3 is 
  low enough to meet the PIC MCUs minimum 
  \$V_{DD}\$ requirements.


The downside of this approach would be that you will lower the voltage even when you are below the 3.6V, so you will not have the full operating range.

Be extremely careful when selecting diodes; \$V_F\$ is not constant with respect to current or temperature.   Here's an example relationship from the Fairchild 1N414 datasheet: 

 

The forward voltage is proportional to the log of the current until about 100mA, when it begins to increase more rapidly due to carrier saturation.  You mention that your device has a minimum current of a few &mu;A, for which you probably paid dearly.  You could increase this quiescent current to a little more than 20 &mu;A with a 220 k&Omega; resistor and put 2 diodes in series to get 450mV drop per diode and a safe output voltage of 3.6V when your batteries are at 4.5V. 

Of course, when the MCU wakes up and draws more current (assuming ~10mA) the voltage drop will go up to about 2⋅700mV=1.4V. Instead of operating from your batteries in their full charge range, you'd only be able to discharge your batteries to 1.8V + 1.4V / 3 = 3.2V / 3 = 1.06 V/cell.  There's some optimization to be had in increasing or decreasing the resistor value and diode count, but it's difficult to get good results as you could with an LDO. 

Conclusion: Don't use this approach unless your application is extremely cost-sensitive!