No, leave it alone.Replace Q2 with a 47K resistor and see what happens??
I went through the old thread, and understand most of what he is doing. Q2 is being used as a zener diode to set the capacitor C1 voltage level at which the fan comes on. 99 other people would use a zener diode, but it's a personal design choice. Could be his version of a watermark, or a trick he learned from a mentor, or ...
The part I don't understand is using a relay to drive a relay. I don't see any reason not to rearrange the circuit around Q1 to eliminate the small relay and have Q1 drive the SSR directly. In fact, I would put the output LED3 in series with the SSR rather than in parallel with it, so it gives a true indication of the output condition. But that's just me.
Also, because he is driving the small relay with an emitter follower rather than a saturated switch, the voltage across the relay coil is going to decrease very slowly. There is an excellent chance that its contacts will "chatter" before opening completely, a not-good thing for whatever is downstream. In your case, the drive circuit inside the SSR might have a little de-bouncing capability to keep the output from producing a small bunch of single-cycle AC bursts to the fan. And, the reed relay coil in his original schemati probably requires more current than the SSR input. To be clear, there is nothing inherently wrong with the double-relay approach, but it adds nothing to the safety of the circuit or any other operational aspects that I can see.
To your original question (no, I didn't forget), there is nothing in your schematic that would make the circuit not work for a while after manufacture, then start working.
ak
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