= Solution
Equality in part (a) requires $a$ to lie in the <span> of the two chosen <unit vectors>. Equivalently, there are real constants $r,s$ such that
$$
\boxed{a_{2i-1}=r,\qquad a_{2i}=s\qquad(1\leq i\leq n).}
$$
One can also see the necessity directly by writing the nonnegative deficit as
$$
n\sum_{i=1}^{2n}a_i^2-\left(\sum_{i=1}^na_{2i}\right)^2-\left(\sum_{i=1}^na_{2i-1}\right)^2
=n\sum_{i=1}^n(a_{2i}-\bar a_e)^2+n\sum_{i=1}^n(a_{2i-1}-\bar a_o)^2,
$$
where $\bar a_e$ and $\bar a_o$ are the means of the even and odd entries. The deficit vanishes exactly when each group is constant; the two group constants need not be equal.
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