Getting everyone to meet everyone else

It can be done in 5 weeks:

  01 02 03 04 05       14 07 04 12 17
  06 07 08 09 10       25 18 22 01 20
  11 12 13 14 15       02 09 03 13 08
  16 17 18 19 20       16 23 06 21 24
  21 22 23 24 25       10 19 05 15 11

  09 19 02 14 17       02 22 23 12 24
  07 21 12 05 03       15 13 14 09 08
  23 25 01 24 11       10 18 11 01 17
  04 15 08 10 22       21 05 16 19 06
  13 16 18 06 20       07 20 03 25 04

  06 16 07 15 12
  14 24 17 10 03
  21 04 02 25 13
  18 20 11 22 05
  09 01 23 08 19

I used a relatively simple computer program to produce the last four seating arrangements. It used a "hill-climbing" optimisation, namely it repeatedly tried to swap to students at random, and if the total number of adjacent pairs increased then keep the swap or else put them back. To produce a seating arrangement, it runs the aforementioned optimisation a few times starting from a random seating, and then picks the best one it found (i.e. the one that introduces the most new pairs given any previous weeks' arrangements already chosen). I then had it produce sets of 5 seating arrangements until it found a set that contained all pairs.

I also set my program to find solutions for $N=6$. It found a 7-week solution:

  01 02 03 04 05 06      12 33 14 22 08 16
  07 08 09 10 11 12      05 23 35 04 06 36
  13 14 15 16 17 18      32 15 19 07 26 09
  19 20 21 22 23 24      01 24 11 28 17 25
  25 26 27 28 29 30      21 13 10 02 20 34
  31 32 33 34 35 36      29 31 30 27 18 03

  01 25 29 27 13 34      22 35 27 13 23 20
  14 03 05 16 02 06      31 14 36 03 12 21
  32 17 19 31 18 15      26 24 11 08 10 18
  35 21 33 09 36 07      15 34 25 07 19 06
  10 08 30 20 12 28      30 28 16 32 29 33
  23 26 11 22 04 24      09 01 04 02 05 17

  14 18 08 20 10 19      16 02 08 10 11 09
  28 05 32 29 01 34      20 34 14 33 21 35
  26 13 09 11 31 12      22 04 29 15 03 05
  03 22 35 23 27 15      36 19 31 07 30 18
  07 24 02 25 04 17      24 17 27 23 32 25
  16 33 36 21 06 30      13 26 01 06 28 12

  19 30 14 03 28 17
  02 12 16 06 31 08
  29 26 35 20 24 27
  04 18 01 05 07 09
  13 33 22 36 34 21
  15 25 10 32 23 11

There is a simple lower bound. There are $N^2$ students, so there are $N^2(N^2-1)/2$ pairs of students that need to sit adjacent to each other at some point. The number of adjacent pairs in one week's seating arrangement is $2N(N-1) + 2(N-1)^2 = 2(N-1)(2N-1)$. Divide them to get a lower bound on the number of weeks. Note that this grows quadratically in $N$.

For $N=5$ we get $300/72=4.167$ so $5$ weeks are necessary.
For $N=6$ we get $630/110=5.727$ so $6$ weeks are necessary.
However, given how hard it was to find a 7-week solution, it is almost surely not possible to attain a 6-week one.

The following are sittings for seven weeks that allow everyone to meet everyone else:

 Week 1            Week 2            Week 3
 [ 0  1  2  3  4]  [18 16 19 15 17]  [ 7  9  5  8  6]
 [ 5  6  7  8  9]  [ 8  6  9  5  7]  [ 2  4  0  3  1]
 [10 11 12 13 14]  [23 21 24 20 22]  [17 19 15 18 16]
 [15 16 17 18 19]  [ 3  1  4  0  2]  [12 14 10 13 11]
 [20 21 22 23 24]  [13 11 14 10 12]  [22 24 20 23 21]

 Week 4            Week 5            Week 6
 [ 8 15  1 20 13]  [20 12  1 19  7]  [22 14  7  0 12]
 [11 22 19  3 18]  [11  9  5 10 21]  [20 16 21  4  3]
 [ 4  6 17 24  5]  [17  8 22 14 23]  [17  1  6  5 23]
 [ 2 14 10  9 16]  [24 13  4  3 15]  [13 18  8  2 15]
 [21 12 23  7  0]  [ 2 18  0 16  6]  [ 9 19 10 11 24]

 Week 7
 [ 1 10 21  7  4]
 [ 3 19 17 18 24]
 [15 11  0 22  5]
 [12 16 23 20 13]
 [19  2  8  6  9]

This is only a partial answer, which shows that the fewest number of weeks in which this can happen is either $5$ or $6$. I suspect the number is $5$, but I do not have a proof.

To see that at least $5$ sittings are required, note that we require a total of $\tbinom{25}{24}=300$ meetings between students. A quick count shows that there are $$\frac12(4\times3+12\times5+9\times8)=72,$$ meetings for each sitting, so we will need at least $5$ weeks for all students to meet.

Here is a series of six sittings in which every student meets every other student, which I have constructed and checked by hand, so there may be a mistake. The time spent constructing this convinces me that it is very likely possible in only five sittings; without (too) much effort, the first five sittings miss only $20$ pairs of students.

01  02  03  04  05
06  07  08  09  10
11  12  13  14  15
16  17  18  19  20
21  22  23  24  25

17  19  22  06  15
08  01  09  24  05
20  18  11  02  14
16  03  10  21  04
13  25  12  23  07

18  25  14  03  24
21  17  06  19  12
02  05  13  04  15
23  20  07  16  01
11  09  10  22  08

09  02  03  07  18
12  25  15  17  04
05  22  23  08  24
11  13  21  01  16
14  20  06  19  10

21  15  13  10  08
24  12  01  25  11
07  14  03  05  04
09  23  16  22  20
17  06  18  02  19

25  07  01  11  15
24  04  19  14  18
13  10  08  05  22
17  02  06  09  23
20  12  21  03  16

Here is an earlier series of seven sittings in which every student meets every other student:

01  02  03  04  05
06  07  08  09  10
11  12  13  14  15
16  17  18  19  20
21  22  23  24  25

17  19  18  15  06
13  01  16  05  24
12  20  03  22  14
02  21  10  25  04
09  11  08  23  07

21  25  10  14  03
18  17  02  08  23
20  04  15  11  05
19  06  01  07  13
22  12  16  09  24

04  10  25  16  20
14  06  13  23  08
03  21  24  11  17
18  01  12  22  07
02  05  19  09  15

06  14  07  21  01
09  17  13  15  25
04  11  03  12  23
20  24  19  02  10
22  08  18  05  16

08  06  03  17  24
22  01  23  20  05
21  14  10  09  25
13  04  07  18  11
02  16  12  19  15

11  12  20  07  05
06  14  01  21  23
18  16  10  19  03
02  24  04  08  09
13  22  15  17  25