I am writing up my comments as an answer. Let $f:\widetilde{X}\to X$ be a minimal desingularization of $X=X_n$. For a general member $\mathcal{C}$ of the family of curves $\mathcal{C}_{p,q}$, let $\widetilde{\mathcal{C}}$ be the strict transform of $\mathcal{C}$ in $\widetilde{X}$.

**Lemma.** The surface $\widetilde{X}$ is a smooth rational surface. The linear system of $\widetilde{\mathcal{C}}$ is basepoint free and big.

**Proof.** This argument is essentially the same as in Mingmin Shen's thesis. By varying $p$ and $q$, it follows that $\widetilde{\mathcal{C}}$ deforms in a family of curves on $\widetilde{X}$ that connect two general points. By the classification of surfaces in characteristic $0$, $\widetilde{X}$ is a rational surface.

Moreover, the normal bundle of $\widetilde{\mathcal{C}}$ in $\widetilde{X}$ is "generically globally generated". By the classification of line bundles on $\mathbb{P}^1$, the normal bundle is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(m)$ for some integer $m>0$. Thus, the normal bundle is globally generated, not just generically globally generated. Therefore, for every point $p\in \widetilde{\mathcal{C}}$, there exists a first-order deformation of $\widetilde{\mathcal{C}}$ that deforms away from $p$. Since the normal bundle has vanishing $h^1$ (by the computation of cohomology of line bundles on $\mathbb{P}^1$), this first-order deformation extends to an honest deformation. Since rational surfaces have $h^1(\widetilde{X},\mathcal{O}_{\widetilde{X}})$ equal to $0$, all of these deformations are in the complete linear system of $\widetilde{\mathcal{C}}$. Thus, the complete linear system is basepoint free and big. **QED.**

**Proposition.** The integer $N$ is bounded above by $n(n+3)/2$.

**Proof.**
Consider the restriction of $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})$ to $\widetilde{\mathcal{C}}$. This is an invertible sheaf on a smooth genus $0$ curve that has degree $n-rm$, where $m$ is the self-intersection number of $\widetilde{\mathcal{C}}$ on $\widetilde{X}$. Notice, $m$ is positive since the linear system is basepoint free and big. Thus, the total degree is negative for $r>n/m$. Since $\widetilde{\mathcal{C}}$ gives a basepoint free linear system, it has nonnegative intersection number with every effective divisor. Thus, $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})$ has only the zero section when $r>n/m$. Therefore we can bound the dimension of the vector space of global sections of $f^*\mathcal{O}(1)$ on $\widetilde{X}$ by the sum over $r=0,\dots,\lfloor n/m \rfloor$ of the dimension of global sections of $f^*\mathcal{O}(1)(-r\widetilde{\mathcal{C}})|_{\widetilde{\mathcal{C}}}$ on $\widetilde{\mathcal{C}}$.

For an invertible sheaf of nonnegative degree $d$ on $\mathbb{P}^1$, the dimension $h^0$ equals $d+1$. Thus, the sum above evaluates to, $$N+1 \leq (\lfloor n/m \rfloor + 1)(n+1 - (m/2) \lfloor n/m \rfloor).$$ This is maximized when $m$ equals $1$. This gives the bound, $$N+1\leq (n+1)(n+2)/2, \ \text{ i.e., } N\leq n(n+3)/2.$$ **QED.**

It is possible to get a classification as in Question 1, but it might not be very explicit. By the computation above, $m\leq n$. Thus, by Mingmin Shen's theorem, the surface $X'$ with hyperplane section $\widetilde{\mathcal{C}}$ is a surface of minimal degree equal to $m$. For each of the finitely many such minimal degree surfaces of degrees $m=1,\dots,n$ in the del Pezzo-Bertini classification, we can list all of the linear systems whose intersection number with $\widetilde{\mathcal{C}}$ has degree $n$. Then we consider sublinear systems of that complete linear system with specified basepoints (basepoints of the linear system on $X'$ give negative self-intersection curves on $X$).

**Edit.** As a corollary of the proof of the proposition, $N$ equals $n(n+3)/2$ only if $X$ is a Veronese $n$-uple surface. Indeed, we need $m$ to equal $1$, so that the surface $X'$ of minimal degree is just $\mathbb{P}^2$ with hyperplane section a line. Moreover, we need the linear system to be the complete linear system of $f^*\mathcal{O}(1)$, or else $N$ would be strictly smaller than $n(n+3)/2$. Thus, there are no base points, and $X$ equals $X'$ embedded by the $n$-uple Veronese map.

anypoint. I edited the question accordingly. $\endgroup$ – Francesco Polizzi Jun 29 '17 at 14:213more comments