14. Let $a, b$ be any real numbers, the solution set of the equation $x^{2}+a|x|+b=0$ in the complex number set $C$ is $\mathrm{A}$. If $|\mathbf{A}|=n$, determine the set $B$ of all possible values of $n$.


14. Let $f(x)=x^{2}+a|x|+b$, then when $n$ is odd, and only when $f(0)=0$, i.e., $b-$
0.
(1) When $n$ is odd, we have $b=0$, i.e., the equation is $x^{2}+a|x|=0$.
If $a>0$, then $A=\{0, \pm ai\}$, hence $n=3$;
If $a=0$, then $A=\{0\}$, hence $n=1$;
If $a<0$, and $a^{2}+4 b>0$, then $\pm \frac{\sqrt{a^{2}+4 b}+a}{2} i$ are the roots of the equation $f(x)=0$, which contradicts $n=0$. If $a^{2}-4 b \geqslant 0$, then $\frac{-a \pm \sqrt{a^{2}-4 b}}{2}$ are the roots of the equation $f(x)=0$, which also contradicts $n=0$.
Discussing when $n \geqslant 8$, the problem does not hold.
If $n \geqslant 8$, then the equation $x^{2}+a x+b=0$ must have two positive roots; the equation $x^{2}-a x+b=0$ must have two negative roots; the equation $x^{2}-a x-b=0$ must have two positive roots; $x^{2}+a x-b=0$ must have two negative roots; by the relationship between roots and coefficients, we know $a>0, b>0 ; a<0, b<0$ must be satisfied simultaneously, which is a contradiction.
Combining (1) and (2), we have $B=\{1,2,3,4,6\}$.