This is one possible solving path. The paired mutually reversing entries mean that some cells are forced to contain the same digit; it may be helpful to map these cells with letters assigned as in the grid shown here. No entry can be a palindrome because its reverse would then be the same, ie a duplicate entry. For any two-digit number ab = 10a + b, the sum of it and its reverse is ab + ba = 10a + b + 10b + a = 11(a + b). Therefore all answers to clues for two-digit entries are multiples of 11, in the range 12 + 21 = 33 to 98 + 89 = 187. For 15ac + 25dn, the only square multiple of 11 in the range is 11² = 121, and the entry pairs adding to that are from the set {29 + 92, 38 + 83, 47 + 74, 56 + 65}.

For 16ac + 3dn, the first Fibonacci numbers are {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …}, of which only 55 is a multiple of 11, so 16ac + 3dn = 55 and the entries are either 14 + 41 (in either order) or 23 + 32. For 9ac + 15dn, the first triangular numbers are {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, …}, of which only 55 and 66 are multiples of 11, but 55 is the answer for 16ac + 3dn, so 9ac + 15dn = 66 with entries from {15 + 51, 24 + 42}. We know 15ac can’t start with 1, so 15dn is in {24, 42, 51}, which means 15ac is in {29, 47, 56} respectively. 9ac can’t start with 3, so 3dn isn’t 23. From the letter map, we now know B is one of [134] and P is one of [976] respectively, ie 24ac is in {19, 37, 46}; in each case the digit sum B + P = 10, so 24ac + 14dn = 11(B + P) = 110 (twice the triangular number 55).

For 13ac + 6dn we have in the grid _[679]_[124] + [124]_[679]_, putting the sum in the range 1601 + 1061 = 2662 to 9994 + 4999 = 14993. The sum of any four-digit number and its reverse is abcd + dcba = (1000a + 100b + 10c + d) + (1000d + 100c + 10b + a) = 1001(a + d) + 110(b + c) = 11(91(a + d) + 10(b + c)), ie a multiple of 11. The only Fibonacci numbers in the required range are {4181, 6765, 10946}, of which the only multiple of 11 is 13ac + 16dn = 6765. To match 1001(a + d) + 110(b + c) = 6765 = 5005 + 1760, we need a + d = 5 and b + c = 16, which rules out 6 as one of the middle digits for 13ac (because the other digit can’t be 10), leaving _7_2 and _9_4. So 13ac + 16dn is one of {3792 + 2973, 1974 + 4791}. In the letter map, this makes K one of {3, 1} and N one of {9, 7} respectively.

For 11ac and 16dn, with digits mapped to letters ELMNL and LNMLE, the sum is the cube of 16ac + 3dn, ie 55³ = 166375. To overflow to six digits, E + L = 15, leaving (166375 − 150015)/10 = 1636 for LMN + NML. Similarly, we need L + N = 16 (ie, 7 and 9 in either order), which leaves (1636 − 1616)/10 = 2 for M + M. Thus 11ac is 87197 if N is 9, or 69179 if N is 7.

From the grid, 8ac = E_B is now either 8_3 or 6_1. If it’s the former, we need 8_3 + 3_8 to be a multiple of 15ac + 25dn = 121. The bounds for that are 803 + 308 = 1111 and 893 + 398 = 1291, always ending in 1; multiples of 121 ending in 1 are {1×121 = 121, 11×121 = 1331, …}, all of which are outside the range. Therefore, 8ac is 6_1, which removes one of the two options for grid letters, leaving B = 1, E = 6, G = 4, H = 2, K = 1, L = 9, N = 7, P = 9, and resolving the entries 9ac = 42, 11ac = 69179, 13ac = 1974, 15ac = 29, 16ac = 41, 24ac = 19, 3dn = 14, 12dn = 97196, 14dn = 91, 15dn = 24, 16dn = 4791 and 25dn = 92. Now 8ac = 6_1 and 6_1 + 1_6 is in the range 601 + 106 = 707 and 691 + 196 = 887, always ending in 7; multiples of 121 ending in 7 are {7×121 = 847, 17×121 = 2057, …} and the only one within the range is 847, making 8ac = 671 and 2dn = 176.

Since (10ac + 19dn)/2 = (23ac + 6dn)/3, 23ac + 6dn must be a multiple of 3, and since both numbers have the same digits, each must itself be a multiple of 3. To make 23ac 9_6 a multiple of 3, the middle digit, J, must be one of {0, 3, 6, 9}. We have 3(10ac + 19dn) = 2(23ac + 6dn), so 0 = 3(604 + 406 + 20×J) − 2(906 + 609 + 20×J) = 20×J, which can only be satisfied by J = 0, making 10ac = 604, 19dn = 406, 23ac = 906 and 6dn = 609.

From the grid, 28ac = 272_ and the nine possibilities for 28ac + 4dn are from 2721 + 1272 = 6435 to 2729 + 9272 = 14443, ie {3993, 4994, 5995, 6996, 7997, 8998, 9999, 11000, 12001}. The nth triangular number is n(n + 1)/2 (because an n×(n + 1) rectangular array of items divided in half by a diagonal line makes two triangles with side length = n), so one way to test whether an integer m is triangular is to find the largest integer n ≤ √(2m) and check whether n(n + 1)/2 = m. Alternatively, m = n(n + 1)/2 means n² + n − 2m = 0, for which the quadratic formula gives the (positive) root n = (−1 + √(1² − 4×1×(−2m)))/2 = (√(8m + 1))/2, which will be an integer (ie m will be triangular) only if (8m + 1) is a square number. Testing the nine values above, only 5595 is triangular, corresponding to 28ac = 2723 and 4dn = 3272. Now 5ac + 22dn = 36F + F63 = 101×F + 423; for that to be a square, (F + 3) must end in one of {0, 1, 4, 5, 6, 9}, so F is one of {7, 8, 1, 2, 3, 6}. Testing the six possibilities {524, 625, 726, 1029, 1130, 1231}, the only square is 625 with F = 2, so 5ac = 362 and 22dn = 263.

The possible values for 26ac + 21dn = 2_6 + 6_2 are {808, 828, 848, 868, 888, 908, 928, 948, 968, 988}, with corresponding digit sums of {16, 18, 20, 22, 24, 17, 19, 21, 23, 25}. For 18ac + 7dn = 9_42 + 24_9 the possible values are {11451, 11561, 11671, 11781, 11891, 12001, 12111, 12221, 12331, 12441}, with corresponding digit sums of {12, 14, 16, 18, 20, 4, 6, 8, 10, 12}. The only shared digit sums are {16, 18, 20}, with the second digit of 18ac, C, being in {2, 3, 4}. We now have 20ac + 10dn = 9CQ96 + 69QC9 = 159105 + 1010×C + 200×Q, in the range 159105 + 1010×2 + 200×0 = 161125 to 159105 + 1010×4 + 200×9 = 164945; it always ends in 5, so if it’s triangular, ie n(n + 1)/2, then n(n + 1) ends in 0 and n ends in one of {0, 4, 5, 9}. The bounds for n are √(2×161125) ≈ 567.7 and √(2×164945) ≈ 574.4 and the possible values are {569, 570, 574} with corresponding triangular numbers {162165, 162735, 165025}. For 569, we need 1010×C + 200×Q = 162165 − 159105 = 3060, but subtracting any of {2020, 3030, 4040} from that doesn’t leave a multiple of 200 for 200×Q. For 574, we need 1010×C + 200×Q = 165025 − 159105 = 5920, from which we can subtract 2020, but the remainder of 3900 isn’t a multiple of 200. That leaves only 570, with 162735 − 159105 = 3630 = 1010×3 + 200×3, ie C = 3, Q = 3, 20ac = 93396, 10dn = 69339, 18ac = 9342 and 7dn = 2439.

The digit sum of 18ac is now 18, so 26ac + 21dn must be 828, with 26ac = 216 and 21dn = 612. 1ac + 17dn is _131 + 131_ and if the missing digit is anything other than 9 the sum would be a palindrome, so 1ac = 9131 and 17dn = 1319. That completes the grid with 27ac = 169 and 1dn = 961, whose sum is indeed the sum of two squares, because both of the entries are squares.