Please see the answer by barak manos for an explanation of what the problem asks. I’ll expand his explanation of the answer a little and show two ways to avoid trial and error, one specific to this problem and one general.
Recall that each participant missed only one problem. Let $x$ be the number of participants who missed a hard problem; each of them solved $9$ easy and $5$ hard problems and therefore solved $9\cdot5=45$ of the hard/easy pairs of problems. Let $y$ be the number of participants who missed an easy problem; each of these participants solved $8$ easy and $6$ hard problems and therefore solved $8\cdot6=48$ of the hard/easy pairs of problems. Altogether the $x+y$ participants solved $45x+48y$ hard/easy pairs, so $45x+48y=459$.
We can divide through by $3$ to reduce this to the equivalent equation $15x+16y=153$. This in turn can be rewritten as $15(x+y)+y=153$, which can be solved by inspection: clearly $153=15\cdot10+3$, so we set $y=3$ and $x+y=10$, finally getting $x=7$. As a check, $45\cdot7+48\cdot3=315+144=459$. Thus, $n=10$, $7$ participants solved all $9$ easy problems and $5$ of the hard problems, and $3$ participants solved all $6$ hard problems and $8$ of the $9$ easy problems.
It was a bit of luck that $153$ could so easily be decomposed in a useful way, so it’s worth pointing out that there are systematic techniques for solving such Diophantine equations.
First note that $15(-1)+16(1)=1$, so $15(-153)+16(153)=153$. That is, $x=-153,y=153$ is a solution to the equation $15x+16y=153$. Next, notice that if $15x_0+16y_0=153$, then
Thus, if $x=x_0,y=y_0$ is a solution, so is $x=x_0+16,y=y_0-15$. In particular, for any integer $k$ we have a solution $x=-153+16k,y=153-15k$. The smallest integer $k$ such that $-153+16k\ge 0$, is $k=10$, giving the solution $x=7,y=3$ that we already found. Any smaller value of $k$ makes $x$ negative, and any larger value makes $y$ negative, so $x=7,y=3$ is the only solution of this kind. And in fact all solutions of $15x+16y=153$ are of this kind (and see also here).
Here I relied on inspection to get one solution ($x=-1,y=1$), but in general the extended Euclidean algorithm can be used to find one mechanically.