Don't put the load-recovery circuit in series with the input.

Quick estimates, assuming the DC converter's 90% efficient and the load-recovery converter is 80% efficient look pretty great. We can turn our 0.6A@5V=3W minimum load into only 0.93W load on the batteries!

But putting the load-recovery circuit in series with the batteries... well, here's a quick thought to get er goin': 80% of 3W is 2.4W. And a 90% efficient DC converter will draw 3.33W to make that 3W [to then make that 2.4W]. 

So now we have a 2.4W source in series with our 40V's worth of batteries... and the sum total is 3.33W, meaning the batteries are only outputting 0.93W! Great! 

But P=VI, and since the sources are in series the current through our 40V batteries [at 0.93W] is the same as the current through our load-recovery voltage-source [at 2.4W]. Again, P=VI, I is constant for both sources, P is comparatively huge from our load-recovery voltage-source, so then V of that voltage-source must also be comparatively huge... to the tune of 103V, on top of the 40V. These DC converters have a wide input range, but nowhere near *that* wide.

So, parallel's probably the way to go, but it seems a bit risky to directly-parallel DC-converter outputs [again, the battery packs are 3.7V with a 5V boost converter]. And then it seems a bit goofy to insert e.g. diodes or resistors... they'll add to the losses we're trying to recover [and likely still, even when the real load is on and the dummy load disabled]. But, may still be worth considering.

...

And, it's kinda an interesting circuit I've come up with to do this... essentially a boost-converter whose feedback node is on the opposite side, measuring/maintaining input *current* rather'n output voltage [or current]. For my first attempt at a homebrew DC-converter, this'd be a doozy!