It is not hard to show that with $36$ people seated at six tables of six places each, there's no way to reseat everybody to a different table each time without pairing some of the same folks a second time. Consider the six people at the first table during the first seating. Where will they go for their second seating? If they must all go to different tables, one of them would have to remain at that first table (pigeonhole principle).
One idea I'm willing to propose involves $35$ people seated at seven tables of five places each. In this scheme we could have (up to) six rotated seatings, with no pair of people meeting more than once, and no person seated at the same table more than once. I note, however, that your preference was to have an even number ($4$ or $6$) of people at each table. What do you think about the proposed type of scheme?
28 People (Seven Tables of Four Each, Three Rounds without Repeats)
Given the desirability of an even number of people at tables (so that games which require a pair of teammates are conducted), I looked for a solution with seven tables and four people at each table (per round). Here's what I found:
First Round
Table 1: A1,B1,C1,D1
Table 2: A2,B2,C2,D2
Table 3: A3,B3,C3,D3
Table 4: A4,B4,C4,D4
Table 5: A5,B5,C5,D5
Table 6: A6,B6,C6,D6
Table 7: A7,B7,C7,D7
Second Round
Table 1: A2,B3,C4,D5
Table 2: A3,B4,C5,D6
Table 3: A4,B5,C6,D7
Table 4: A5,B6,C7,D1
Table 5: A6,B7,C1,D2
Table 6: A7,B1,C2,D3
Table 7: A1,B2,C3,D4
Third Round
Table 1: A3,B5,C7,D2
Table 2: A4,B6,C1,D3
Table 3: A5,B7,C2,D4
Table 4: A6,B1,C3,D5
Table 5: A7,B2,C4,D6
Table 6: A1,B3,C5,D7
Table 7: A2,B4,C6,D1
It is fairly easy to verify that all twenty-eight people are seated in each round, and that the three seatings at each table are free of repeats. Less easy to spot is whether there are any pairs of people who meet more than once. The design is based on a rotation of a certain kind, so that it actually suffices to check any one person (for duplicate meetings) to be sure that no one meets the same person twice. Here are the seatings for A1 (and you can see there are no repetitions):
Round 1: A1,B1,C1,D1 (Table 1)
Round 2: A1,B2,C3,D4 (Table 7)
Round 3: A1,B3,C5,D7 (Table 6)
It follows that each participant will meet nine other people during this networking event.
Best Answer
This is the Social Golfer Problem. You're looking for Resolvable Steiner Quadruple Systems. The highest known solution I know about is for 32 golfers playing in foursomes for 10 days.
The names of what actually needs to be looked for are somewhat scattered. For example, 36 golfers in foursomes can play for 11 days. This is a 4-RGDD of type $2^{18}$.
Theoretically, the 100 golfers in your group could be split into foursomes 33 (1 person + 3x33) times, but that would be a perfect covering, and those are pretty well known. This isn't a perfect covering listed in the Handbook of Combinatorial Designs, so it's likely something tricky, or nonperfect and trickier.
The La Jolla Covering Repository stops at 99 points. There is a C(99,4,2) = 817 design... if this was resolvable (splittable into groups covering 1-100), 817/25 = 32.65, so maybe you could get 32 splits.
Maybe the Design of the Century is resolvable. In that case, you've got a perfect solution. That paper has various references for this particular design. In any case, you'll find an answer within combinatorial methods far faster than with any kind of brute force approach.