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Explanation:

Let there be n players in each of these 4 groups.
∴ There are total 4n players.

Let the groups be called G1, G2, G3 & G4.

Now, a player from G1, say A, plays with other (n - 1) players from G1 exactly once.
Hence, he plays (n -1) games.

A will play with each of n players of G2 twice, hence total 2n games.

A will play with each of n players of G3 thrice, hence total 3n games.

A will play with each of n players of G4 four times, hence total 4n games.

∴ A will play a total of (n - 1) + 2n + 3n + 4n = 10n - 1 games.

Now, each of the 4n players will play (10n - 1) games.
∴ There will be total 4n × (10n - 1)/2 games [/2 because each game is counted twice]
⇒ There are total 2n(10n - 1) games

⇒ 2n(10n - 1) > 1000
⇒ n(10n - 1) > 500

The least possible value of n satisfying the above inequality is 8.

Hence, 8.

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