I can confirm the 20-ticket-set will provide (at least) one 3-OK when drawing 6 balls out of 24.
The attachements show some analyses i made of this set.
You can see it favors certain combinations much and ticket 1 7 8 14 16 18 would be best to be drawn .
You can also see it favors some number over others much (number 20 used 7 times, while 7-10-17-22 are used just 3 times) and has a rather strange distribution of combinations of 2-balls occurring (in fact just 138 [being exactly 50%] of the 276 possible actually do occur).
Certain 3-OK , 4-OK and 5-OK combinations occur very often:
3-OK
4-OK
5-OK
1 8 14 ---> 4
1 8 14 16 ---> 4
1 8 14 16 18 ---> 4
1 8 16 ---> 4
1 8 14 18 ---> 4
1 8 18 ---> 4
1 8 16 18 ---> 4
1 14 16 ---> 4
1 14 16 18 ---> 4
1 14 18 ---> 4
8 14 16 18 ---> 4
1 16 18 ---> 4
8 14 16 ---> 4
8 14 18 ---> 4
8 16 18 ---> 4
14 16 18 ---> 4
I've been surprised by that 116 tickets answer provided by GillesD as it showed ALL of the 2024 possible combinations of 3-OK to be covered and also a good (allthough not 'perfect') distribution of 2-balls occurences (in my Excelsheet you can see for the 20-tickets set in the attachements).
To me that 116-sets solution looked like overkill (being confirmed by the 20-ticket solution). I myself have been working on a 24ticket solution with equal occurence for each ball and each combination of 3-OK occuring JUST once while all 2-balls-combination are covered as well (most of them just once, some 2 times) as I believe(d ??) in a uniform distribution of the things mentioned for an optimal result.
I wonder (have my big doubts) if the 163-tickets for C(49,6,3) is optimal.
Both along the lines of my investigations and the C(24,6,3)=20 I believe there will be a MUCH lower coverage possible (I expect below 50 tickets).
Allthough it being nice to have (at least) 1 3-OK within 20 tickets I wonder how that solution will allow to say something like "garanteed X tickets will have a 3-OK if you play 24 times those 20-tickets moving each ball 1 position at a time". Would X be 24 or would it be allowed to to say X being a value over 24 and how much over 24??
In this i'm not sure if a solution using a bit more tickets but with much better distribution of combinations and balls wouldn't be beter as such a solution has far less sets being covered JUST once.
Also I wonder how that 20-ticket solution will work out for winning larger prices as it covers just 282 out of 10626 combinations of 4-OK and 117 out of 42504 of 5-OK. Even if playing those tickets 24 times (like suggested above) coverage of those combinations will be small.
Buying 20 tickets will still be a very small chance of winning so its use actually is 'reducing losses' Actually if a ticket will cost 1 euro and a 3-OK will provide about 10 euro it means you can have 2 chances for something above 3-Ok for the price of one, so "you can have one chance for free".
It would be interesting to see a C(24,6,4) and/or a C(49,6,4) similar like the C(24,6,3)=20 .
Remark : I tried to add 2 (small) JPG files but it seems not posisble
as the OP asked for a minimum number. 20 is certainly in reach of more people than 116. With wheels, you decrease the likelihood of larger prizes in return for a minimum (worst case 
