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.