= Solution
\b[Yes; without <self-loops>, such a nontrivial <simple cooperative game> has exactly one positive <edge>, of weight $1$.] For each pair,
$$
w_{ij}=v(\{i,j\})\in\{0,1\},
$$
because singleton and loop contributions are zero. Summing all <edge weights> gives $v(N)=1$, so precisely one unordered pair $\{a,b\}$ has weight $1$, and all other pairs have weight zero.
Thus the <winning coalitions> are exactly those containing both $a$ and $b$. For any $0\le\lambda\le1$, set
$$
\boxed{x_a=\lambda,\quad x_b=1-\lambda,\quad x_i=0\quad(i\notin\{a,b\}).}
$$
This <payoff> <vector> is efficient, since $\sum_i x_i=v(N)=1$. Every <winning coalition> receives $1$ and every losing <coalition> receives a nonnegative total, so all inequalities $x(S)\ge v(S)$ hold. Hence it belongs to the <core of a cooperative game>.
In fact these are all <core> allocations. Singleton conditions require $x_i\ge0$, and the winning pair requires $x_a+x_b\ge1$; efficiency then forces equality and zero <payoffs> outside the pair. The endpoints are both <veto players>, consistent with the general characterization of the <core of a simple cooperative game>.
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