This is a harder problem than the 13 team one. I believe I have a workable solution below, where again there is the odd empty slot in the schedule and again you will need to leave out the last four rounds. The problem is that team D plays twice on day 1 and four times on day 2. An alternative might be to have one extra empty slot on day 2, and remove the empty slot on day 3 - meaning 2 teams would play three times on day 3.
Schedule:
(J K) (A H) (G I)
(---) (C E) (B D)
(H I) (F K) (A J)
(C G) (B E) (---)
(H K) (A D) (F J)
(E I) (B G) (C F)
------------------
(F H) (E J) (D G)
(A K) (C I) (B H)
(D F) (---) (E G)
(B J) (I K) (A C)
(D E) (G H) (A F)
(C J) (D K) (B I)
------------------
(A B) (D J) (F I)
(G K) (---) (E H)
(A I) (B F) (C D)
(G J) (C H) (E K)
------------------
(B C) (D I) (H J)
(E F) (A G) (C K)
(D H) (I J) (---)
(A E) (F G) (B K)
Team frequency in each round
A B C D E F G H I J K
R001 1 0 0 0 0 0 1 1 1 1 1
R002 0 1 1 1 1 0 0 0 0 0 0
R003 1 0 0 0 0 1 0 1 1 1 1
R004 0 1 1 0 1 0 1 0 0 0 0
R005 1 0 0 1 0 1 0 1 0 1 1
R006 0 1 1 0 1 1 1 0 1 0 0
--------------------------
R007 0 0 0 1 1 1 1 1 0 1 0
R008 1 1 1 0 0 0 0 1 1 0 1
R009 0 0 0 1 1 1 1 0 0 0 0
R010 1 1 1 0 0 0 0 0 1 1 1
R011 1 0 0 1 1 1 1 1 0 0 0
R012 0 1 1 1 0 0 0 0 1 1 1
--------------------------
R013 1 1 0 1 0 1 0 0 1 1 0
R014 0 0 0 0 1 0 1 1 0 0 1
R015 1 1 1 1 0 1 0 0 1 0 0
R016 0 0 1 0 1 0 1 1 0 1 1
--------------------------
R017 0 1 1 1 0 0 0 1 1 1 0
R018 1 0 1 0 1 1 1 0 0 0 1
R019 0 0 0 1 0 0 0 1 1 1 0
R020 1 1 0 0 1 1 1 0 0 0 1