= Solution
Choose an orientation for each curve and let $c_i=[C_i]\in H_1(X;\mathbb Q)$. The <algebraic intersection number of curves on an oriented surface> gives a nondegenerate <alternating bilinear form> on the $2g$-dimensional space $H_1(X;\mathbb Q)$, by <Poincare duality>. Distinct curves are disjoint, so their intersection numbers are zero; each self-pairing is zero by alternation. Thus the span $W=\operatorname{span}\{c_1,\ldots,c_n\}$ is an <isotropic subspace of a symplectic vector space>. The assumed <linear independence> gives $\dim W=n$.
Nondegeneracy identifies $H_1(X;\mathbb Q)$ with its dual. Restricting functionals to $W$ is surjective, so the <symplectic orthogonal complement> has dimension $\dim W^\perp=2g-n$. Isotropy says $W\subseteq W^\perp$, giving $n\le2g-n$. Hence
$$
\boxed{n\le g.}
$$
The bound is sharp: one meridian on each handle of a genus-$g$ surface gives $g$ pairwise disjoint curves with independent <homology (mathematics)> classes. This is an algebraic bound on independent classes, not on the number of disjoint separating or mutually homologous curves.
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