There are $\binom{m}2$ pairs of students, and each project takes care of $\binom{n}2$ pairs. Therefore, at least $$\binom{m}2\Big/\binom{n}2=\frac{m(m-1)}{n(n-1)}$$ days are necessary.
If this ratio is an integer, then a schedule which attains is called a Steiner system $S(2,n,m)$. We know that $S(2,q,q^2+q)$ and $S(2,q+1,q^2+q+1)$ exist for any prime power $q$. Furthermore, $S(2,3,6k+1)$ and $S(2,3,6k+3)$ exist for all $k\ge 1$. See here for more details on what is known about the existence of Steiner systems.
We can get an upper bound on the length of an optimal schedule using the probabilistic method. Suppose you randomly choose a $d$ day schedule. The probability a particular pair of people is not chosen is $\Big(1-\binom{m-2}{n}/\binom{m}n\Big)^d=\Big(1-\frac{(m-n)(m-n-1)}{m(m-1)}\Big)^d$. Therefore, the probability that at least one pair of students is not chosen is at most
$$
\binom{m}2\Big(1-\frac{(m-n)(m-n-1)}{m(m-1)}\Big)^d
$$
If you choose $d$ large enough so that quantity is less than $1$, then it means there is a positive probability that a random $d$ day schedule works, so that in particular it can be done in $d$ days. Being very rough, if we let $x=\frac{m(m-1)}{(m-n)(m-n-1)}$, then this is $\binom{m}2 (1-\frac1{x})^{x\cdot d/x}\approx \binom{m}2e^{-d/x}$, so this is satisfied when $$d\gtrsim x\log\binom{m}2\approx \frac{m(m-1)}{(m-n)(m-n-1)}\times 2\log m.$$
The problem with your second working is that it leads to duplicates. To understand it, when you have chosen $9$ students of $14$ and arranged them in between $10$ students who cannot sit together, say there is a student A who is in the group of $9$. Now you are left with group of $5$ students. Say there is a student B in this group of $5$. Now you seat B in one of the twenty available slots.
But you also have arrangements where B is chosen to be in the first group of $9$ and A is in the second group of $5$. Now when you try and seat A as part of second group, it leads to duplicate arrangements.
If you want to seat $10$ students first and then seat $14$ students, here is one way to count arrangements -
First use stars and bars to count number of possible arrangements of seats. There are $11$ slots available after seating $10$ students. $9$ slots in between cannot be empty and that leaves only $5$ seats that must go to one of the $11$ slots. That can be done in $\displaystyle {15 \choose 5}$ ways.
Now that we have $14$ seats, all is left is to seat the students, which can be done in $14!$ ways.
So total number of permissible arrangements = $\displaystyle 10! \cdot {15 \choose 5} \cdot 14!$
Best Answer
It can be done in 5 weeks:
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:
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.