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Did anyone else feel there was an ambiguity raised by the penultimate clue? I reckoned that there were two possible numbers that would work in the top LH corner. One had the advantage of making the relevant grid entries into squares, but reading the preamble carefully that wasn't required. Maybe one of the options uses a number that appears elsewhere but I couldn't spot it.

It’s true, cruncher, that 1d and 27a individually need not be squares. But in my understanding, the first two digits of 27a are fixed by 16d and 19d, and thus the second and third digits of 1d are also fixed. This means the only variable is the third digit of 27a, I.e first digit of 1d. But only one value satisfies the non-palindrome specification of the first clue.

Did I misunderstand you, or does that make sense?

Dodgepot - you are quite right. I've dismantled this grid so many times that I was being dim. I had noticed that 7 in the top corner would give a valid answer for the squares but had then assumed the non-palindrome would involve adding two 7s together for its first two digits - which of course it doesn't because it adds to its reverse. Doh. Glad that one is over. I got stuck on earlier attempts because I was trying to make the digit sum clues work around the digits in their entries rather than the digits of their clue answers
. Took me 4 days to spot that...

Took me about 4 days to spot exactly that as well cruncher. Finally finished.

It interests me that so many solvers fell into the same trap, as I did at first, of considering the sum of the digit sums of the two individual answers, rather than the digit sum of the answers added together.

Well, that was fun. Finally crawled over the line after just two restarts, the first occasioned by missing a critical triangular number out of my crib table. Grrr.

Still, very neat, and I particularly like the elegant way that the middle figure conundrum for 10a/26d and 23a/6d was resolved.

Nice to have completed this one after being defeated by the last two numericals.

Having come rather late to this puzzle I have enjoyed tackling most of it but am now well and truly stuck ! Is the "same digit sum" a prime number ? I have the same number for the third and fourth cells of 10 Down; is that correct? The other 4-cell triangular number is baffling me somewhat: are cells 2 and 3 three times a square number ? Any help or hints and/or confirmation of my points will be very much appreciated !!

Same digit sum - neither 18ac nor 7d is prime and their digit sum isn't prime either
10d 3rd/4th digit are the same
4d 2nd/3rd digits are 3 times a square but not so in 28ac

Hi jordan thanks to your valuable pointers plus confirmations, and my re-reading Posts 43 to 45, I have finally limped over the line !

The challenge with these puzzles is in identifying the path of least resistance (as opposed to the simplest looking clues) and when you do they can be solved much quicker (that being a relative term!). Starting with clues that give multiple solutions greater than four can lead to endless cycles and inevitable mistakes.
Generally speaking clues involving numbers raised to higher powers than 2, or solutions in sequences that have a decreasing frequency of occurence versus order of magnitude are a good starting point. This one become a pleasure after overcoming the tendency to confuse the solutions with the mathematical entities.