Let's go for an explicit analytic construction of the set of all the
solutions to your problem.
Basic notation: $e_K := \text{ column vector of }K\text{
}1'$s. $\langle x, y \rangle$ denotes the inner product between two
vectors $x$ and $y$. For example $\langle e_K, x\rangle$ simply
amounts to summing the components of $x$.
Recall the following definitions, to be used without further explanation.
Preimage: Given abstract sets $X$, $Y$, $Z \subseteq Y$.,
the preimage of $Z$ under $f$ is defined by $f^{-1}Z := \{x \in X |
f(x) \in Z\}$. This has nothing to do with function inversion.
Convex hull: Given a subset $C$ of $\mathbb{R}^K$, its convex
hull is defined by $\textit{conv }C :=$ smallest convex subset of
$\mathbb{R}^K$ containing $C$.
Indicator function: Given a subset $C$ of $\mathbb{R}^K$ its
indicator function $i_C:\mathbb{R}^K \rightarrow (-\infty, +\infty]$
is defined by $i_C(x) = 0$ if $x \in C$; otherwise $i_C(x) = +\infty$.
Simplex: The $K$-dimensional simplex is defined by $\Delta_K :=
\{x \in \mathbb{R}^K|x\ge 0, \langle e_K, x\rangle =
1\}$. Equivalently, $\Delta_K = \{\text{rows of the }K \times K\text{ identity matrix }I_K\}$. Each row of $I_K$ is a vertex of $\Delta_K$.
Faces of a simplex: Given a subset of indices $I \subseteq
\{1,2,...,K\}$, the face of $\Delta_K$ spanned by $I$, denoted
$F_K(I)$, is defined to be the convex hull of those rows of $I_k$
indexed by $I$. Trivially, $\Delta_K$ is a face of
itself. Geometrically, $F_K(I)$ is a $\#I$-dimensional simplex
whose vertices are those vertices of $\Delta_K$ which labelled by $I$.
Let $f:\mathbb{R}^K \rightarrow (-\infty,+\infty]$ be an extended
real-value function.
Subdifferential: $\partial f(x) := \{g \in \mathbb{R}^K| f(z) \ge
f(s) + \langle g, z - x\rangle, \forall z \in \mathbb{R}^K\}$.
Fenchel-Legengre transform: $f^*(x) := \underset{z \in \mathbb{R}^K}{\text{max
}}\langle x, z\rangle - f(z)$.
The optimal value of your objective under the given constraints can be
conveniently written as
\begin{eqnarray}
v = \underset{x \in \mathbb{R}^K}{\text{min }}g(x) + f(Dx),
\end{eqnarray}
where $D := a^T \in \mathbb{R}^{1 \times K}$, $g := i_{\Delta_K}$ and
$f:s \mapsto \frac{1}{2}s^2$. One recognizes the above problem as the
primal form of the saddle-point problem
\begin{eqnarray*}
\underset{x \in \mathbb{R}^K}{\text{min }}\underset{s \in
\mathbb{R}}{\text{max }}\langle s, Dx\rangle + g(x) - f^*(s)
\end{eqnarray*}
Given a dual solution $\hat{s} \in \mathbb{R}$, the entire set
of primal solutions $\hat{X}$ is given by (one can check that
sufficient conditions that warrant the strong Fenchel duality
Theorem)
\begin{eqnarray*}
\hat{X} = \partial g^*(-D^T\hat{s}) \cap D^{-1}\partial
f^*(\hat{s}).
\end{eqnarray*}
For your problem, one easily computes, $g^*(y) = \underset{x \in
\Delta_K}{\text{max }}\langle x, y\rangle$, and so by the
Bertsekas-Danskin theorem (see Proposition A.22 of Bertsekas' PhD thesis) for subdifferentials, we have
\begin{eqnarray*}
\begin{split}
\partial
g^*(-D^T\hat{s}) &= \partial g^*(-\hat{s}a) = ... \text{( some computations masked )} \\
&= F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a
minimal component of }\hat{s}a\}\right).
\end{split}
\end{eqnarray*}
Also $\partial f^*(\hat{s}) = \{\hat{s}\}$, and so $D^{-1}\partial
f^*(\hat{s}) = \{x \in \mathbb{R}^K|\langle a, x\rangle =
\hat{s}\}$. Thus the set of
all primal solutions is
\begin{eqnarray*}
\hat{X} = F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a
minimal component of }\hat{s}a\}\right) \cap \{x
\in \mathbb{R}^K|\langle a, x\rangle = \hat{s}\}.
\end{eqnarray*}
It remains now to find a dual solution. Define $\alpha := \underset{1
\le i \le K}{\text{min }}a_i \le \beta := \underset{1
\le i \le K}{\text{max }}a_i$. One computes
\begin{eqnarray*}
\begin{split}
v &= \underset{x \in \Delta_K}{\text{min }}\underset{s \in
\mathbb{R}}{\text{max }}s\langle a, x\rangle
-\frac{1}{2}s^2 = \underset{s, \lambda \in
\mathbb{R}}{\text{max }}\underset{x \ge 0}{\text{min }}\langle sa +
\lambda e_K, x\rangle - \frac{1}{2}s^2 - \lambda\\
&=\underset{s, \lambda \in
\mathbb{R}}{\text{max }}-\frac{1}{2}s^2 - \lambda\text{ subject
to }sa + \lambda e_K \ge 0\\
&=-\underset{s, \lambda \in
\mathbb{R}}{\text{min }}\frac{1}{2}s^2 + \lambda\text{ subject
to }\lambda \ge -\underset{1 \le i \le K}{\text{min }}sa_i\\
&= -\underset{s \in
\mathbb{R}}{\text{min }}\frac{1}{2}s^2 -\underset{1 \le i \le
K}{\text{min }}sa_i
= -\frac{1}{2}\underset{s \in
\mathbb{R}}{\text{min }}\begin{cases}(s - \alpha)^2 -
\alpha^2, &\mbox{ if }s \ge 0,\\(s -
\beta)^2 - \beta^2, &\mbox{ otherwise}\end{cases}
\end{split}
\end{eqnarray*}
Thus
\begin{eqnarray*}
\hat{s} = \begin{cases}\beta, &\mbox{ if }\beta < 0,\\0,
&\mbox{ if }\alpha \le 0 \le \beta,\\\alpha, &\mbox{ if }\alpha >
0.\end{cases}
\end{eqnarray*}
Therefore the set of all solutions to the original / primal problem is
\begin{eqnarray*}
\hat{X} = \begin{cases}F_K\left(\{1 \le i \le K | a_i =
\beta\}\right),
&\mbox{ if }\beta < 0,\\
\Delta_K \cap \{x \in \mathbb{R}^K| \langle a, x\rangle = 0\},
&\mbox{ if } \alpha \le 0 \le \beta,\\
F_K\left(\{1 \le i \le K | a_i = \alpha\}\right),
&\mbox{ if }\alpha > 0,\end{cases}
\end{eqnarray*}
Now that you have the hammer, find the nails...
Case (1): $\beta < 0$. Choose any $k \in \{1,2,...,K\}$ such that
$a_k = \beta$, and take $\hat{x} = k$th row of $I_K$.
Case (2a): $a_k = 0$ for some $k$. Take $\hat{x} = k$th row of $I_K$.
Case (2b): $\alpha < 0 < \beta$, and $a_k \ne 0 \forall k$.
Choose $k_1, k_2 \in \{1,2,...,k\}$ such that $a_{k_1} < 0 < a_{k_2}$,
and take
$\hat{x}_k = \frac{1}{{a_{k_2} - a_{k_1}}}\begin{cases}a_{k_2},
&\mbox{ if }k = k_1,\\-a_{k_1},&\mbox{ if
}k = k_2,\\0, &\mbox{ otherwise.}\end{cases}$
Case (3): $\alpha > 0$. Choose any $k \in \{1,2,...,K\}$ such that
$a_k = \alpha$, and take $\hat{x} = k$th row of $I_K$.
