= Solution
\b[True for a smooth complex projective surface], the usual manifold convention in this question. Choose an <ample divisor> $H$. The auxiliary pencil-existence theorem states that, for all sufficiently large $m$, a general two-dimensional space of sections of $\mathcal O_X(mH)$ defines a <Lefschetz pencil>: its base points are transverse and finite, its general fibre is a smooth connected curve, and each singular fibre has exactly one ordinary node, with distinct critical values. High powers separate the required jets, so a generic pencil has precisely these singularities.
Choose such an $m$ divisible by four, and write $D=mH$. The <adjunction formula> for a smooth connected curve of class $D$ on a smooth projective surface is $2g-2=D\cdot(D+K_X)$. Consequently
$$
g=1+\frac{m^2H^2+mK_X\cdot H}{2}.
$$
Writing $m=4\ell$ gives $g=1+8\ell^2H^2+2\ell K_X\cdot H$, an odd integer. Thus \b[a sufficiently positive four-divisible polarization gives an odd-genus <Lefschetz pencil>]. The surface is taken to be smooth; a singular projective variety need not support a <Lefschetz pencil> in this smooth-manifold sense.
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