$13 \cdot 3$ If $a, c, d$ are integers, $b$ is a positive integer, and they satisfy $a+b=c$, $b+c=d, \quad c+d=a$. Then, the maximum value of $a+b+c+d$ is
(A) -1 .
(B) -5 .
(C) 0 .
(D) 1 .
(China Junior High School Mathematics League, 1991)
[Sol] $a+b=c$ and $c+d=a$ are added, getting $a+b+c+d=a+c$.
So $b+d=0$, which means $b=-d$. Substituting $b=-d$ into $b+c=d$, we get $c=2d$.
Therefore, $a=c+d=2d+d=3d$.
Thus, $a+b+c+d=a+c=3d+2d=5d=-5b$.
Since $b$ is a positive integer, the minimum value of $b$ is 1.
Therefore, the maximum value of $a+b+c+d$ is -5.
Hence, the answer is $(B)$.