Group 1:
- Number of students $=N$
- Average fees paid $=17,500$
- Total fees paid by Group $\(1=N \times 17,500=17500 N\)$

Group 2:
- Number of students $=N-1$
- Average fees paid $=22,500$
- Total fees paid by Group $\(2=(N-1) \times 22,500=22500(N-1)\)$

Quantity A: Average fees paid by $2 N-1$ students
The total number of students in both groups combined is $N+(N-1)=2 N-1$.
The total fees paid by all $2 N-1$ students is the sum of fees from Group 1 and Group 2:
Total fees $=17500 N+22500(N-1)$
Total fees $=17500 N+22500 N-22500$
Total fees $=40000 N-22500$
Average fees paid by $2 N-1$ students $\(=\frac{\text { Total fees }}{\text { Total number of students }}\)$
Average fees $\(=\frac{40000 N-22500}{2 N-1}\)$
Now, let's compare this with Quantity B, which is 20000.

We need to compare $\(\frac{40000 N-22500}{2 N-1}\)$ with 20000.
Let's test some values for $N$. Since $N$ is a number of students, $N$ must be a positive integer.
Also, $N-1$ must be at least 1 , so $\(N \geq 2\)$.

Case 1: Let $N=2$
- Number of students in Group $1=2$
- Total fees Group $\(1=2 \times 17500=35000\)$
- Number of students in Group $2=2-1=1$
- Total fees Group $\(2=1 \times 22500=22500\)$
- Total students $=2+1=3$
- Total fees $=35000+22500=57500$
- Quantity A = Average fees = $\(\frac{57500}{3}=19166.67\)$ (approximately)
- Quantity B $=20000$

In this case, Quantity A (19166.67) is less than Quantity B (20000).
Case 2: Let $N=3$
- Number of students in Group $1=3$
- Total fees Group $\(1=3 \times 17500=52500\)$
- Number of students in Group $2=3-1=2$
- Total fees Group $\(2=2 \times 22500=45000\)$
- Total students $=3+2=5$
- Total fees $=52500+45000=97500$
- Quantity A = Average fees = $\(\frac{97500}{5}=19500\)$
- Quantity B $=20000$

In this case, Quantity A (19500) is less than Quantity B (20000).

It appears Quantity A is always less than Quantity B. Let's try to prove this algebraically.
We want to compare $\(\frac{40000 N-22500}{2 N-1}\)$ with 20000.
Multiply both sides by $(2 N-1)$ (which is positive since $\(N \geq 2\)$ ):
Compare $40000 N-22500$ with $20000(2 N-1)$
Compare $40000 N-22500$ with $40000 N-20000$
Subtract $40000 N$ from both sides:
Compare -22500 with -20000
Since $-22500<-20000$,
This means Quantity A is always less than Quantity B.
The final answer is Quantity B is greater.