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* = 10*a* + *b*,
the sum of it and its reverse is *ab* + *ba*
= 10*a* + *b* + 10*b* + *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 11^{2} = 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*
= (1000*a* + 100*b* + 10*c* + *d*)
+ (1000*d* + 100*c* + 10*b* + *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 55^{3} = 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 *n*th 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* ≤ √(2*m*)
and check whether *n*(*n* + 1)/2 = *m*.
Alternatively, *m* = *n*(*n* + 1)/2 means
*n*^{2} + *n* − 2*m* = 0,
for which the quadratic formula gives the (positive) root
*n* = (−1 + √(1^{2} − 4×1×(−2*m*)))/2
= (√(8*m* + 1))/2,
which will be an integer (ie *m* will be triangular)
only if (8*m* + 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.