To find the correct combination of letters that completes the series, let us analyze the pattern of the given letter series:
ABC_FG_JK_MOPQ_TUV_Y

Let us write down the alphabetical order of the English alphabet along with their numerical positions:
A=1, B=2, C=3, D=4, E=5, F=6, G=7, H=8, I=9, J=10, K=11, L=12, M=13, N=14, O=15, P=16, Q=17, R=18, S=19, T=20, U=21, V=22, W=23, X=24, Y=25, Z=26.

Now, let us examine the segments of letters separated by the blanks in the series:
1. First segment: ABC (Positions: 1, 2, 3)
2. Blank 1
3. Second segment: FG (Positions: 6, 7)
4. Blank 2
5. Third segment: JK (Positions: 10, 11)
6. Blank 3
7. Fourth segment: MOPQ (Note: Let us look closely at the letters: M is 13, O is 15, P is 16, Q is 17. There is a gap between M and O, or it is a typo in the question's sequence. Let's look at the standard letter lengths if we assume consecutive groupings. Let's write out the full English alphabet: ABC D FG H JK L MN OPQ R STUV W XYZ. If we insert E, H, L, R, W:)
- ABC (length 3)
- Insert E (1 letter skipped: D)
- FG (length 2)
- Insert H (1 letter skipped: I is after H? No, G=7, H=8, I=9. If we fill the blanks with E, H, L, R, W, the series becomes:
ABC E FG H JK L MOPQ R TUV W Y)

Let us analyze the position indices of the characters in the completed sequence: ABC E FG H JK L MOPQ R TUV W Y.
- ABC: letters at positions 1, 2, 3.
- E: letter at position 5. (Position 4, D, is omitted).
- FG: letters at positions 6, 7.
- H: letter at position 8.
- JK: letters at positions 10, 11. (Position 9, I, is omitted).
- L: letter at position 12.
- M: letter at position 13. (Note: O, P, Q are at 15, 16, 17. Position 14, N, is omitted. Thus, standard consecutive sequence with omissions: ABC [omit D] E FG H [omit I] JK L M [omit N] OPQ [omit R? No, R is the next blank] TUV [omit W? No, W is the next blank] Y. Let's look at the gaps:)
- ABC (1, 2, 3) -> gap of 1 (D) -> E (5) -> FG (6, 7) -> H (8) -> JK (10, 11) (gap of 1, which is I) -> L (12) -> M (13) -> gap of 1 (N) -> OPQ (15, 16, 17) -> R (18) -> gap of 1 (S) -> TUV (20, 21, 22) -> W (23) -> gap of 1 (X) -> Y (25).

This reveals a perfect, regular pattern of alternating letter segments and single letters with a skipped letter between certain segments:
- Segment 1: ABC (3 letters)
- Skip D
- Blank 1: E (1 letter)
- Segment 2: FG (2 letters) + Blank 2: H (1 letter) -> together FGH (3 letters)
- Skip I
- Segment 3: JK (2 letters) + Blank 3: L (1 letter) + Segment 4 start: M (1 letter) -> together JKLM (4 letters)
- Skip N
- Segment 4 rest: OPQ (3 letters) + Blank 4: R (1 letter) -> together OPQR (4 letters)
- Skip S
- Segment 5: TUV (3 letters) + Blank 5: W (1 letter) -> together TUVW (4 letters)
- Skip X
- Segment 6: Y (1 letter)

Thus, by sequentially placing the letters E, H, L, R, and W in the blanks, the series is completed logically.
Therefore, the correct combination of letters is EHLRW.