A mere 6 plies are needed to reach a position in which White has a dual free #4. For any length record, as long as there is at least one unique mating line, it qualifies. This is in spirit as some checkmating problems are created with a line of "best defense." This involves one side leaving the other with only a single possible continuation in a vain hope.
[FEN ""]
1. e3 b6 2. Bc4 Bb7 3. Qf3 d6 {Now the helping play is over.} 4. Qxf7+ Kd7 5. Qf5+ e6 6. Bxe6+ Kc6 (6... Ke7 7. Qf7#) 7. Qd5#
A #5 can be done in 6 plies.
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1. e4 e5 2. Bc4 f5 3. Qe2 h6 {Now the helping play is over.} 4. Qh5+ g6 5. Qxg6+ Ke7 6. Qf7+ Kd6 7. Qd5+ Ke7 8. Qxe5#
A #6 can be done in 5 plies; see Noam Elkies's answer.
A #7 can be done in 8 plies.
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1. e4 e5 2. Bc4 d5 3. Qh5 Ke7 4. f4 Kf6 {Now the helping play is over.} 5. Qg5+ Ke6 6. Qxe5+ Kd7 7. Qxd5+ Ke7 8. Qxf7+ Kd6 9. Qd5+ Ke7 10. Qe5+ Be6 (10... Kd7 11. Qe6#) 11. Qxe6#
A #8 can be done in 10 plies.
[FEN ""]
1. e3 e5 2. Qh5 Ke7 3. Bc4 Kf6 4. c3 g6 5. d3 a6 {Now the helping play is over.} 6. Qf3+ Kg7 7. Qxf7+ Kh6 8. e4+ Qg5 9. Bxg5+ (9... Kxg5 10. h4+ Kh6 11. Qxf8+ Kh5 12. g4+ Kxg4 13. Qf3#) Kh5 10. g4+ Kxg4 11. h4 Nf6 12. Qxf6 Kh5 13. Qf3#
A #9 can be done in 6 plies.
[FEN ""]
1. f4 e5 2. Kf2 Qh4+ 3. Ke3 Bc5+ {Now the helping play is over.} 4. Kd3 e4+ 5. Kc3 Qf6+ 6. d4 exd3+ 7. Kd2 Qxf4+ 8. e3 Qxe3+ 9. Kc3 d2+ 10. Bd3 Qd4+ 11. Kxd2 Qf4+ 12. Kc3 (12. Ke2 Qf2#) (12. Ke1 Qf2#) 12... Qb4#
A #10 can be done in 5 plies.
[FEN ""]
1. d4 e5 2. Kd2 Bb4+ 3. Ke3 {Now the helping play is over.} Qg5+ 4. Kd3 e4+ 5. Kc4 b5+ 6. Kxb4 a5+ 7. Kc3 b4+ 8. Kb3 a4+ 9. Kc4 Ba6+ 10. Kxb4 Nc6+ 11. Kc3 Qa5+ 12. b4 Qxb4#