Denote $T_k$ the team placed on the $k$-th place and $p_k$ the number of points of this team, $k\in\{1,2,3,4,5\}$. Let $P$ be the total number of points awarded to the five teams. If $p_k$ are consecutive numbers, then $P=5p_3$. Caution: this condition is necessary, but not sufficient. The teams play $\binom{5}{2}=10$ matches. Denote $a$ the number of matches which end with the victory of one of the tams, $b$ the number of draws with goals and $c$ the number of draws without goals. Results: $a+b+c=10$. In a match with victory, the two teams obtain together $5$ points. In a draw with goals, the two teams obtain together $4$ points. In a draw without goals, the two teams obtain together $2$ points. The total number of points is $P=5a+4b+2c$. $P=5p_3\Longrightarrow 5a+4b+2c=5p_3\Longrightarrow 5|2b+c$. If a match ends with victory, is scored at least $1$ goal (score $1-0$). In the case of draw with goals, are scored at least $2$ goals (score $1-1$). Hence, the least number of goals scored in the tournament is $G=a+2b$. We will determine the possible values of $b,c$ such that $b+c\le10$ and $5|2b+c$. Automatically result the values of $a=10-b-c$ and $G=a+2b$. The possible configurations of types of matches and minimum number of goals $(a,b,c,G)$ are: $(10,0,0,10); (5,0,5,5); (0,0,10,0); (6,1,3,8); (1,1,8,3)$; $(7,2,1,11); (2,2,6,6); (3,3,4,9); (4,4,2,12)$; $(5,5,0,15); (0,5,5,10); (1,6,3,13); (2,7,1,16)$. We study the configurations in the ascending order of $G$: a) $(a,b,c,G)=(0,0,10,0)$ implies each team obtains $4$ points, hence $p_k$ are not consecutive numbers. b) $(a,b,c,G)=(1,1,8,3)$. Results: $P=25; p_1=7; p_2=6;p_3=5;p_4=4;p_5=3$. Each team plays $4$ matches. One of the teams $T_1, T_2, T_3$ obtains the single victory and this team doesn't lose any match. Results for this team $p_k\ge 5+1+1+1=8>7=p_1$, contradiction. Hence, this configurations is not possible. c) $(a,b,c,G)=(5,0,5,5)$. Results: $P=35; p_1=9; p_2=8;p_3=7;p_4=6;p_5=5$. But $T_1$ cannot obtain $9$ points only from victories and draws without goals (for $2$ victories: $p_1\ge 2\cdot5=10>9$; for $1$ victory: $p_1\le 5+1+1+1=8<9)$. d) $(a,b,c,G)=(2,2,6,6)$. Results: $P=30; p_1=8; p_2=7;p_3=6;p_4=5;p_5=4$. Exists a configuration which satisfies the conditions: We associate to each team the configuration (wins, draws with goals, draws without goals, losses): $T_1: (1,0,3,0)$; $T_2: (1,0,2,1)$; $T_3: (0,2,2,0)$; $T_4: (0,1,3,0)$; $T_5: (0,1,2,1)$. In this case we obtain the minimum number of goals scored in the tournament: $G=a+2b=2+2\cdot2=6$. The scores for which we obtain this minimum are: $T_1:T_2\; 1-0; T_1:T_3\; 0-0; T_1:T_4\; 0-0; T_1:T_5\; 0-0; T_2:T_3\; 0-0$; $T_2:T_4\; 0-0; T_2:T_5\; 1-0; T_3:T_4\; 1-1; T_3:T_5\; 1-1; T_4:T_5 \;0-0$.