Group them into $6$ pairs of two schoolchildren. We will show that each pair knows exactly four songs and that no two pairs know a common song. This will clearly be sufficient. Indeed, consider any five pairs knowing $20$ songs. Since any four of them know $16$ songs, each of the five pairs knows at least $4$ songs which none of the other four pairs know. But then these $4 \cdot 5$ "single songs" already contribute $20$ so each pair knows exactly four songs and no two of the five pairs know a common song. Since the five were chosen arbitrarily, this establishes our claim.