Assume that
$$\sum_i^n{|a^{(1)}_i-b^{(1)}_i|} \le \sum_i^n{|a^{(2)}_i-b^{(2)}_i|}$$
where for $j=1,2$, there is $a_i^{(j)}\ge0$, $\sum_{i}^{n}a_i^{(j)}=1$, $b_i^{(j)}\ge0$ and $\sum_{i}^{n}b_i^{(j)}=1$,
dose the following hold,
$$\sum_i^n{|a^{(1)}_i-b^{(1)}_i|^2} \le \sum_i^n{|a^{(2)}_i-b^{(2)}_i|^2}$$
Probably not, let $b^{(1)} = b^{(2)} = (1/4, 1/4, 1/4, 1/4)$, $a^{(1)} = b^{(1)} + (-\epsilon, -2\epsilon, -3\epsilon, 6\epsilon)$ and $a^{(2)} = b^{(2)} + (-3\epsilon, -3\epsilon, 3\epsilon, 3\epsilon)$ for some small $\epsilon$, then $$\sum_i |a_i^{(1)} - b_i^{(1)}| = 12\epsilon = \sum_i |a_i^{(2)} -b_i^{(2)}|,$$ but $$\sum_i |a_i^{(1)} - b_i^{(1)}|^2 = (1^2 + 2^2 + 3^2 + 6^2)\epsilon^2 > 4\times 3^2\epsilon = \sum_i |a_i^{(2)} -b_i^{(2)}|$$
More generally, your claim would imply that $$\sum_i |a_i^{(1)} - b_i^{(1)}|= \sum_i |a_i^{(2)} - b_i^{(2)}|$$ implies $$\sum_i |a_i^{(1)} - b_i^{(1)}|^2 = \sum_i |a_i^{(2)} - b_i^{(2)}|^2$$ which immediately seems wrong.