Step 1: 

It is given that average number of coins per sack in the boxes are all distinct integers, so, each box will have average of 1, 2, 3, 4, 5, 6, 7, 8 and 9 coins per sack in any order. 

Therefore, each row and column will have (9*10)/2 = 45 coins in total, since each row and column has same number of coins. 

Also, since average number of coins per sack is an integer value therefore, the sum of coins in each sack in a box should be divisible by 3. 

Average of ‘1’ is only possible for the combination of (1, 1, 1). 

This combination is possible for boxes in 3rd column - 3rd row because this satisfies 2 conditions that minimum value is 1 and median is also 1. 

Average ‘9’ is only possible for the combination of (9, 9, 9). 

This combination is possible in 2nd row-3rd column (Box in 3rd row 1st column has median 8, therefore, it cannot have an average value of 9). 

Box in 3rd row - 1st column satisfies condition (iii) only therefore, the only combination of coins that satisfies the mentioned condition is (7, 8, 9). Box in 3rd row/2nd column must satisfy conditions (i) and (iii). 

So, minimum and maximum number of coins in a bag are 1, and 9 respectively. Hence, possible combinations are (1, 5, 9) or (1, 8, 9) but only (1, 8, 9) will give the sum of 45 for Row-3 

Step 2: 

If average number of coins in a box is ‘2’ then, total sum of coins is suppose to be 6 which means each sack must have less than 5 coins and that is only possible for box in 2nd row - 2nd column. 

This box satisfy only one condition and that must be (I).

Hence, the only possible combination is (1, 2, 3). 

[We take (1, 1, 4) then two given conditions will be satisfied which contradicts.]