In a chess tournament, there are four groups, each containing an equal number of players. Each player plays
• against every other player belonging to one's own group exactly once;
• against each player belonging to one of the remaining three groups exactly twice;
• against each player belonging to one of the remaining two groups exactly three times; and
• against each player belonging to the remaining group exactly four times.
If there are more than 1000 matches being played in the tournament, the minimum possible number of players in each group is_______.

Let's there are 4 teams named A,B,C,D . And let the no of players in each team be equal to 'n' . 

For team A : 
1. against every player of A exactly once =  $$n_{C_2}$$
2. against every player of team B exactly twice = (n)(2n) 
3. against every player of team C exactly thrice = (n)(3n) 
4. against every player of team D exactly 4 times = n(4n) 

For team B :
1.  against every player of B exactly once = $$n_{C_2}$$
3.  against every player of team D exactly thrice = (n)(3n) 
4.  against every player of team C exactly 4 times = n(4n)
against every player of team A exactly twice is n(2n) but which is already counted in A . Hence to avoid recounting lets take it = 0.

For team C :
1. against every player of C exactly once =  $$n_{C_2}$$
2. against every player of team D exactly twice = (n)(2n)
Rest of the thrice and four times with A & B respectively are already counted above.

For team D :
1. against every player of D exactly once = $$n_{C_2}$$
Rest all in team D is already counted above . 

Hence, total no. of matches = 4($$n_{C_2}$$) + $$2n^2(2+3+4)$$
                                           = $$2n(n-1)+18n^2$$ > 1000

Hence, the least integral value of 'n' is 8.