The puzzle’s title is an anagram of FOURTHSANDIVEYW, the distinct letters of the puzzle’s number, 4542.

This is one possible solution path. All the grid entries are of two or three digits, so they’re in the range 10 to 999. In that range, no names of numbers contain any of BGJKMPQZ, so the maximum possible letter sum for any entry is the sum of the remaining letters, ie Σ(ADEFGHILNORSTUVWXY) = 253, and any three-digit entry that is the clue for another entry must start with 1 or 2.

A is a two-digit number whose name has 6 letters,
ie one of {11, 12, 20, 30, 80, 90}.
It can’t end with 0 (which would make entry a start with 0),
so A is either 11 with a letter sum of Σ(ELVN) = 5 + 12 + 22 + 14 = 53,
or 12 with a sum of Σ(TWELV) = 20 + 23 + 5 + 12 + 22 = 82.
A’s clue says its letter sum is n,
and n’s clue says it’s a 10-letter number,
which rules out “eighty-two”,
so **n = 53** and **A = 11**.

The clue for n is G, so G is the letter sum for 53,
ie Σ(FITYHRE) = **G = 91**.
Then the clue for G is d, so Σ(NIETYO) = **d = 88**,
and the clue for d is F, so Σ(EIGHTY) = **F = 74**.
Now h is a three-digit number starting with 4,
and its name has 11 letters, so it must be **h = 400**
(with the letter sum Σ(FOURHNDE) = 91 = G).

B is the clue for N, so B starts with 1 or 2,
which means b is in the range [10, 29];
but b’s name has 9 letters, so it’s one of {17, 21, 22, 26}.
The letter sum of F = 74 is Σ(SEVNTYFOUR) = 165
and from F’s clue we know that equals k − b,
so k = b + 165 and is one of {182, 186, 187, 191}.
Of those, only “one hundred and eighty-seven” has 24 letters,
so **k = 187** and **b = 22**.
The letter sum of b is Σ(TWENYO) = 102,
which agrees with its clue of A + G = 11 + 91 = 102.
From clue K we have B − b = d + G,
so B = d + G + b = 88 + 91 + 22, ie **B = 201**.
Then B’s letter sum gives Σ(TWOHUNDREA) = **e = 129**.

The letter sum of k is Σ(ONEHUDRAIGTYSV) = **H = 188**.
The letter sum of H then gives Σ(ONEHUDRAIGTY) = **E = 147**,
which in turn gives Σ(ONEHUDRAFTYSV) = **f = 178**,
whose sum Σ(ONEHUDRASVTYIG) = 188 matches H.

From the grid, g is 41_ and the clue says its name has 23 letters,
so it can only be **g = 417**.
Its letter sum is Σ(FOURHNDEASVT) = 153 = (H + D)/2,
and as we know H = 188, that gives **D = 118**.
Then D’s letter sum is Σ(ONEHUDRAIGT) = 122 = M − H,
which makes **M = 310**.

From the grid we have C = 1_2,
and the clue says its length is 23,
which can only be “one hundred and seventy-two”,
so **C = 172**.
From the grid we have m = 81_,
and the clue says its length is 22,
so it’s either 815 with a letter sum of Σ(EIGHTUNDRAF) = 113
or 816 with a letter sum of Σ(EIGHTUNDRASX) = 150.
Its clue says the letter sum is C − b = 172 − 22 = 150,
so **m = 816**.

For p we have 3_ and its length is 9,
so p is one of {31, 32, 36};
its letter sum is D = 118,
which matches only Σ(THIRYWO), so **p = 32**.
N is now 7_2, with a length of 23,
ie one of {742, 752, 762};
its letter sum is B = 201,
which matches only Σ(SEVNHUDRAFOTYW), so **N = 742**.

Finally, P is 6_ with a length of 10,
ie one of {63, 67, 68};
its letter sum is c = 117,
which matches only Σ(SIXTYEGH),
so **P = 68** and the grid is complete.