Lower bound for a combinatorial problem ($N$ students taking $n$ exams)

A better-than-linear lower bound for $N$ is $\frac{1}{4} (n \log_2 n + n)$.

To see why, let's look at what you can do when $n$ is a power of $2$.

For $n = 2^1 = 2$, we may take $N_2=3$. Our $N=3$ students are called $a,b,c$, and in our $n=2$ exams, the top two students are:

Exam 1: $\ a \ \ b$

Exam 2: $\ a \ \ c$

It is clear that student $a$ will be selected in the first round, and then either $b$ or $c$ will be selected next, depending on the ordering of the exams.

For $n = 2^2 = 4$, we may take $N_4=8$. Our $N=8$ students are called $a,b,c,d,e,f,g,h$, and in our $n=4$ exams, the top four students are:

Exam 1: $\ a \ \ b \ \ c \ \ e$

Exam 2: $\ a \ \ b \ \ c \ \ f$

Exam 3: $\ a \ \ b \ \ d \ \ g$

Exam 4: $\ a \ \ b \ \ d \ \ h$

It is clear that student $a$ and $b$ will be selected in the first two rounds. No one else is guaranteed selection, but it is not difficult to check that any given student is selected for some ordering of the four exams.

For $n = 2^3 = 8$, we may take $N_8=20$. Our $N=20$ students are called $a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t$, and in our $n=8$ exams, the top eight students are:

Exam 1: $\ a \ \ b \ \ c \ \ d \ \ \ e \ \ g \ \ i \ \ m$

Exam 2: $\ a \ \ b \ \ c \ \ d \ \ \ e \ \ g \ \ i \ \ n$

Exam 3: $\ a \ \ b \ \ c \ \ d \ \ \ e \ \ g \ \ j \ \ o$

Exam 4: $\ a \ \ b \ \ c \ \ d \ \ \ e \ \ g \ \ j \ \ p$

Exam 5: $\ a \ \ b \ \ c \ \ d \ \ f \ \ h \ \ k \ \ q$

Exam 6: $\ a \ \ b \ \ c \ \ d \ \ f \ \ h \ \ k \ \ r$

Exam 7: $\ a \ \ b \ \ c \ \ d \ \ f \ \ h \ \ l \ \ s$

Exam 8: $\ a \ \ b \ \ c \ \ d \ \ f \ \ h \ \ l \ \ t$

This time students $a - d$ are guaranteed selection, but no one else is. Again, it is not difficult to check that any given student is selected for some ordering of the four exams.

Without going to the trouble of writing down a formal recurrence for this pattern, you can check that if $n = 2^k$ then a solution like this leads to a situation where $$N = 2^{k-1} \cdot 1 + 2^{k-2} \cdot 2 + 2^{k-3} \cdot 4 + 2 \cdot 2^{k-2} + 1 \cdot 2^{k-1} + 2^k = 2^{k-1}(k + 2).$$ Putting everything in terms of $n$, this is $N_{2^k} = \frac{n}{2}(\log_2 n + 2)$.

For some $n$ that is not a power of $2$, we can (crudely) just round down to the nearest power of $2$ for the desired solution: pick $k$ such that $2^k < n \leq 2^{k+1}$ and then observe that $$N_n \geq N_{2^k} \geq (k+2)2^{k-1} = \frac{1}{4}(2^{k+1} \cdot k + 2 \cdot 2^{k+1}) = \frac{1}{4}(2^{k+1}(k+1) + 2^{m+1}) \geq \frac{1}{4}(n \log_2 n+n)$$ as claimed.

My initial argument was erroneous. I'm rewriting everything completely.

Denote by $f(n)$ the maximal value of $N$ for that value of $n$. I claim that $$ f(k+\ell)\geq f(k)+f(\ell)+k \qquad\text{for all $k\leq \ell$.} \qquad\qquad(*) $$ In a standard way, this shows that $f(n)\geq 1+\frac12n\log_2 n$.

To prove $(*)$, assume that we have $k+\ell$ exams. Take a set $X$ of $f(k)$ students hopeful for the first $k$ exams and hopeless for the last $\ell$, and take a set $Y$ of $f(\ell)$ students performing in the opposite way. Finally, take a set $Z$ of $k$ students, each of which wins his own exam in the first group and his own exam in the last group and not attenfing the others.

Now, if the first $k$ exams are taken into account first, then $Z$ win their exams and become out of competition, so $Y$ are still hopeful. Similarly, $X$ stay hopeful if the last $\ell$ exams are taken into account first.