When seeking answers in the future, it is best to ask in advance what scope of knowledge to use to solve them, so as not to waste time and so on. I'm wasting my energy to solve it for everyone's convenience:

Primary school method:

9 1 population1-1/4-n/5 = 3/4-n/5 = (15-4n)/20.

Then the total number of people is 91(15-n)/20) =1820/(15-4n).

1820= 1*2*2*5*7* 13

15-4N must be the product of 1, 2, 2, 5, 7, 13 and 1 to 6 factors.

N = 1, 15-4n = 1 1, which does not meet the requirements.

N=2, 15-4N=7, which meets the requirements.

N=3, 15-4N=3, which is not satisfactory.

N=4, 15-4N=- 1, which is unreasonable.

1820/( 15-4N)= 1820/7 = 260

The number of participants is 260.

Let the total number of people be X.

x-9 1=x/4+Nx/5

x = 9 1/( 1- 1/4-N/5)= 1820/( 15-4N)

1820=2^2*5* 13*7

Therefore, n = 1, 15-4n = 1 1 cannot be divisible by 1820.

N=2, 15-4N=7, divisible by 1820.

N=3, 15-4N=3, which cannot be divisible by 1820.

N=4, 15-4N=- 1, which is unreasonable.

So only N=2 meets the requirements.

Then x =1820/(15-4n) =1820/7 = 260.

The number of participants is 260.