There are multiple solutions ofcourse, find below the way I thought through this;

Total milk is 3321, so each son should get 369.

If you write the full sequence (1 to 81) in a straight line and then add the extreme numbers in pairs (1+81, 2+80, 3+79 and so on), each pair will add up to 82. 4 such pairs will add upto 328. So lets give 4 such pairs to each son, so 8 cows to each son. Each of them will have 328 with them. The order looks like below;

Son 1 – 01, 02, 03, 04 and 81, 80, 79, 78 – the first 4 and the last 4 cows
Son 2 – 05, 06, 07, 08 and 77, 76, 75, 74 – the next 4 cows from either end
Son 3 – 09, 10, 11, 12 and 73, 72, 71, 70 – and so on..
Son 4 – 13, 14, 15, 16 and 69, 68, 67, 66
Son 5 – 17, 18, 19, 20 and 65, 64, 63, 62
Son 6 – 21, 22, 23, 24 and 61, 60, 59, 58
Son 7 – 25, 26, 27, 28 and 57, 56, 55, 54
Son 8 – 29, 30, 31, 32 and 53, 52, 51, 50
Son 9 – 33, 34, 35, 36 and 49, 48, 47, 46

Now there are 9 cows remaining 37 to 45.

To get to 369, each of them need 41 more. 41 is the 5th in sequence. So lets give 41 to the 5th son. Now he has 328+41 = 369. He is sorted.

Now consider the remaining sons in pairs (Son 1 and Son 2), (Son 3 and Son 4), (Son 6 and Son 7), (Son 8 and Son 9)
If you give 37 to son 1, then he is 4 short. So swap a pair of cows between the two sons that has a difference of 4 (basically any cow), Say they swap their first cows

So at this stage,
Son 1 – 05, 02, 03, 04 and 81, 80, 79, 78 = 332+37 = 369
Son 2 – 01, 06, 07, 08 and 77, 76, 75, 74 = 324. So he needs a cow that covers for the 41 he was short in the first place and the 4 he gave to Son 1  – so the only option is 45. So this pair is sorted.

Take the next pair. Each of them are 41 short, but we only have cows which are +/-3, +/-2 and +/-1 away from 41. For this pair, lets do with +/-3. So give 38 to Son 3. He will now be 3 short. So swap a cow with a difference of 3. There are 6 possibilities. Lets go with 10 from Son 1 is swapped with 13 from Son 2.
Son 3 – 09, 13, 11, 12 and 73, 72, 71, 70 = 331+38 = 369
Son 4 – 10, 14, 15, 16 and 69, 68, 67, 66 = 325+44 = 369

For the next pair, they are again 41 short, but we only have cows which are +/-2 and +/-1 away from 41. If we do +/-2 for this pair, then give 39 to Son 6. He will now be 2 short. Swap a cow with Son 7 with a difference of 2. So Son 6 will now gain 2 and be equal to 369. Son 7 will now need to make up for the 41 he needed anyway and the 2 he lost to Son 6. The only option is 43.
Son 6 – 21, 22, 25, 24 and 61, 60, 59, 58 = 330+39 = 369
Son 7 – 23, 26, 27, 28 and 57, 56, 55, 54 = 326+43 = 369

For the final pair, we only have cows 40 and 42 remaining (+/-1 away from 41). Give 40 to Son 8. He will now be 1 short. Swap a cow with Son 9 with a difference of 1. So Son 8 will now gain 1 and be equal to 369. Son 9 will now need to make up for the 41 he needed anyway and the 1 he lost to Son 8. The only option is 42.
Son 8 – 29, 30, 31, 33 and 53, 52, 51, 50 = 329+40 = 369
Son 9 – 32, 34, 35, 36 and 49, 48, 47, 46 = 327+42 = 369

As you can see, in this method alone, there are many combinations possible.

Hope you had the patience to read through this 🙂

Cheers

A farmer has 81 cows numbered 1 to 81.
No 1 cow gives 1litre milk … no 2 gives 2litre…… no 81 gives 81 litre per day.
1litre + 2litre + 3litre + .. + 81litre = 3321 litre
The farmer has 9 sons.
Each one of the 9 sons should take 9 cows each, which yield a total of
one-ninth of 3321 litre milk, i.e., 369 kg of milk.
How he can distribute?
He can distribute as shown below:
………………………..Cow Number…………………
Son No. 1 … 1 11 21 31 41 51 61 71 81
Son No. 2 … 2 12 22 32 42 52 62 72 73
Son No. 3 … 3 13 23 33 43 53 63 64 74
Son No. 4 … 4 14 24 34 44 54 55 65 75
Son No. 5 … 5 15 25 35 45 46 56 66 76
Son No. 6 … 6 16 26 36 37 47 57 67 77
Son No. 7 … 7 17 27 28 38 48 58 58 78
Son No. 8 … 8 18 19 29 39 49 59 69 79
Son No. 9 … 9 10 20 30 40 50 60 70 60

I hope it’s answer should be 20.
Probably can be solved using Theory of Magic squares
see these links to explore.. how Magic squares are created
https://www.math.hmc.edu/funfacts/ffiles/10001.4-8.shtml

this one to see.. that Each magic sqaure can produce 8 distinct magic squares
https://en.wikipedia.org/wiki/Magic_square