Modulo 9/3 Tetramagic Series Lemma
If
  A, D, B, E, C, F are the number of
  1, 2, 4, 5, 7, 8 (mod 9) entries, respectively,
  in a tetramagic series.
and thus,
  (A+B+C) is the number of entries of 1 (mod 3) and
  (D+E+F) is the number of entries of 2 (mod 3)
then
  (A+B+C) ≡ 4(S2 + S3) + S4 + S3 (mod 9) and
  (D+E+F) ≡ 4(S2 + S3) + S4      (mod 9).

Proof

[1] A + 7B + 4C + 4D + 7E +  F ≡ S2 (mod 9)
[2] A +  B +  C + 8D + 8E + 8F ≡ S3 (mod 9)
[3] A + 4B + 7C + 7D + 4E +  F ≡ S4 (mod 9)

Subtract [2] from [1].
[4]     6B + 3C + 5D + 8E + 2F ≡ S2 - S3 (mod 9)

Subtract [2] from [3].
[5]     3B + 6C + 8D + 5E + 2F ≡ S4 - S3 (mod 9)

Multiply [5] by 2.
[6]    6B + 3C + 7D +  E + 4F ≡ 2(S4 - S3) (mod 9)

Subtract [4] from [6].
[7]  2D + 2E + 2F ≡ 2S4 - S3 - S2 (mod 9)

Multiply [7] by 5.
[8] (D+E+F) ≡ 4(S2 + S3) + S4 (mod 9).

Substitute [8] into [2].
[9] (A+B+C) ≡ 4(S2 + S3) + S4 + S3 (mod 9).

Equations [8] and [9] give the desired result.