We number chess players in accordance with their power: $P_1, P_2, P_3, P_4, P_5, P_6$ (if the $i$'th player is stronger than the $j$'th one, then $i < j$). Suppose, contrary to our claim, that the tournament can be organized. Let chess player $P_i$ get $a_i$ points ($i = 1, \dots, 6$). By conditions, $a_i - a_{i+1} \ge 0.5$ for $i = 1, \dots, 5$. In particular
$$a_i - a_{i+3} = (a_i - a_{i+1}) + (a_{i+1} - a_{i+2}) + (a_{i+2} - a_{i+3}) \ge 0.5 + 0.5 + 0.5 = 1.5. \quad (1)$$
Let $A = \{P_1, P_2, P_3\}$, $B = \{P_4, P_5, P_6\}$, $S_A = a_1 + a_2 + a_3$, $S_B = a_4 + a_5 + a_6$. It is easy to see that $S_A + S_B = 15$. Moreover $S_A - S_B \ge 4.5$. Indeed
$$S_A - S_B = (a_1 + a_2 + a_3) - (a_4 + a_5 + a_6) = (a_1 - a_4) + (a_2 - a_5) + (a_3 - a_6) \ge 1.5 + 1.5 + 1.5 = 4.5. \quad (2)$$
Let $S_Y(X)$ denote the total number of points got by the chess players of group X in the games with the chess players of group Y (here $X, Y \in \{A, B\}$).
It is evident that $S_A(A) = S_B(B) = 3$. It follows that $S_B(A) - S_A(B) = (S_A(A) + S_B(A)) - (S_A(B) + S_B(B)) = S_A - S_B$. The total number of the games between the players of A and B is equal to $3 \cdot 3 = 9$, so $9 \cdot 1 = 9$ points are distributed in these games. So for $S_A - S_B \ge 4.5$ it is necessary to have $S_B(A) \ge 7$. Otherwise, $S_B(A) \le 6.5$ gives $S_A(B) = 9 - S_B(A) \ge 2.5$, which implies $S_A - S_B \le 6.5 - 2.5 = 4$, contrary to (2).
For $S_B(A) \ge 7$ it is necessary to use the fair board at least 5 times, otherwise, since only using the fair board gives 1 point to a stronger player, $S_B(A) \le 4 \cdot 1 + 5 \cdot 0.5 = 6.5$, which is impossible.
Since the fair board is used exactly 5 times, the players from A must use the peaceful board for the remaining $4 = 9-5$ games with the players from B to obtain $S_B(A) \ge 7 = 5+4 \cdot 0.5$.
Therefore the peaceful board is used at most 1 time for the games between the players of the same group. It follows that $a_6 \ge 1 + 0.5 = 1.5$. Then
$$15 = S_A + S_B = a_6 + a_5 + a_4 + a_3 + a_2 + a_1 \ge 1.5 + 2 + 2.5 + 3 + 3.5 + 4 = 16.5,$$
a contradiction. Hence it is impossible to organize the tournament satisfying the problem conditions.