B. Given that $a$ is an integer. If there exist two unequal integers $x, y$ satisfying the system of equations $\left\{\begin{array}{l}x-y^{2}=a, \\ y-x^{2}=a,\end{array}\right.$ and $|x| \leqslant 10$, then the minimum value of $a$ is . $\qquad$
B. -111 .

From $x-y^{2}=y-x^{2}$
$$
\Rightarrow(x-y)(x+y+1)=0 \text {. }
$$

Since $x \neq y$, then $x+y+1=0$.
Substituting $y=-(x+1)$ into the original equation, we get
$$
a=-\left(x^{2}+x+1\right)=-\left(x+\frac{1}{2}\right)^{2}-\frac{3}{4} \text {, }
$$

where $-10 \leqslant x \leqslant 10$.
When $x=10$, $a$ has the minimum value
$$
-\left(10^{2}+10+1\right)=-111
$$