= Solution
Let $B(X,Y)=\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y)$ be the <Killing form>. A real <semisimple Lie algebra> has nondegenerate $B$, and its invariance gives $B([X,Y],Z)=B(X,[Y,Z])$. Translating $B$ defines a <bi-invariant pseudo-Riemannian metric> $g=B$ on the <Lie group>.
For <left-invariant vector fields>, all derivatives of their pairwise inner products vanish. The <Koszul formula> reduces to
$$
2B(\nabla_XY,Z)=B([X,Y],Z)-B([Y,Z],X)+B([Z,X],Y)=B([X,Y],Z).
$$
Nondegeneracy gives $\nabla_XY=[X,Y]/2$. With $R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z$, the <Jacobi identity> now gives
$$
R(X,Y)Z=-\tfrac14[[X,Y],Z].
$$
Taking the trace of $X\mapsto R(X,Y)Z=-\tfrac14\operatorname{ad}Z\operatorname{ad}Y(X)$ yields the <Ricci tensor>
$$
\boxed{\operatorname{Ric}(Y,Z)=-\tfrac14 B(Y,Z)=-\tfrac14 g(Y,Z).}
$$
Thus the <Killing-form Einstein metric> makes the group an <Einstein manifold>, allowing an indefinite metric. Its <scalar curvature> in this convention is $-\dim G/4$.
For a connected compact <semisimple Lie group>, $B$ is negative definite: the adjoint operators are skew-adjoint for an invariant positive inner product, so $B(X,X)=\operatorname{tr}((\operatorname{ad}X)^2)<0$ for $X\ne0$. The conventional positive <Riemannian metric> is instead $g_+=-B$. Its <Ricci tensor> is $g_+/4$, and its <sectional curvature> is
$$
K_{g_+}(X,Y)=\frac{\|[X,Y]\|_{g_+}^2}{4(\|X\|_{g_+}^2\|Y\|_{g_+}^2-\langle X,Y\rangle_{g_+}^2)}\geq0.
$$
It can vanish on commuting two-planes; positivity is not automatic in every direction. With the literal negative metric $B$, the corresponding sectional-curvature signs are reversed.
For a noncompact real semisimple group, a Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$ has $B$ negative on $\mathfrak k$ and positive on $\mathfrak p$. The <Killing metric> is indefinite and cannot be made positive by an overall sign. Its Einstein constant remains $-1/4$ for $g=B$, but that does not assert a uniform sign for sectional curvature of all nondegenerate planes. For connected semisimple groups, negative-definite $B$ characterizes the compact case. Disconnected groups require a separate condition on the number of components before compactness follows.
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