Let $\rm V$ be the $2 \times n$ matrix whose $i$-th column is vector $\rm v_i$. We have the Boolean optimization problem in $\mathrm z \in {0,1}^n$
$$\min_{\mathrm z \in {0,1}^n} \| \mathrm V \mathrm z + \mathrm w \|_2^2$$
Let
$$\rm z = \frac 12 (x + 1_n)$$
where $\mathrm x \in {\pm 1}^n$. 945
Hence, we have an instance of the Boolean least-squares (BLS) problem
$$\min_{\mathrm x \in {\pm 1}^n} \| \mathrm A \mathrm x - \mathrm b \|_2^2$$
where
$$\rm A = \frac 12 V \qquad \qquad \qquad b = - \left( \frac 12 V 1_n + w \right)$$
Since ${\pm 1}$ is the solution set of the quadratic equation $x_i^2 = 1$, we have the following (non-convex) quadratically constrained quadratic program (QCQP) in $\mathrm x \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \| \mathrm A \mathrm x - \mathrm b \|_2^2\ \text{subject to} & x_i^2 = 1 \quad \forall i \in {1,2,\dots,n}\end{array}$$
Exhaustive evaluation of the objective function at all $2^n$ points is only feasible for small $n$.
SDP relaxation
Note that
$$\begin{array}{rl} \| \mathrm A \mathrm x - \mathrm b \|_2^2 &= (\mathrm A \mathrm x - \mathrm b)^{\top} (\mathrm A \mathrm x - \mathrm b)\ &= \mbox{tr} \left( (\mathrm A \mathrm x - \mathrm b) (\mathrm A \mathrm x - \mathrm b)^{\top} \right)\ &= \mbox{tr} \left( \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x\ 1\end{bmatrix} \begin{bmatrix} \mathrm x\ 1\end{bmatrix}^{\top} \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix}^{\top} \right)\ &= \mbox{tr} \left( \begin{bmatrix} \mathrm A^{\top}\ -\mathrm b^{\top}\end{bmatrix} \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x\ 1\end{bmatrix} \begin{bmatrix} \mathrm x^{\top} & 1\end{bmatrix} \right)\ &= \mbox{tr} \left( \begin{bmatrix} \,\,\,\, \mathrm A^{\top} \mathrm A & -\mathrm A^{\top}\mathrm b\ -\mathrm b^{\top}\mathrm A & \,\,\,\, \mathrm b^{\top}\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x \mathrm x^{\top} & \mathrm x\ \mathrm x^{\top} & 1\end{bmatrix} \right)\end{array}$$
Since $\mathrm x \in {\pm 1}^n$, all $n$ entries on the main diagonal of $\mathrm x \mathrm x^{\top}$ are equal to $1$. Thus,
$$\begin{bmatrix} \mathrm x \mathrm x^{\top} & \mathrm x\ \mathrm x^{\top} & 1\end{bmatrix}$$
is symmetric, positive semidefinite and has only ones on its main diagonal, i.e., it is a correlation matrix. Its rank is $1$. 262