Schematic locus of algebraic spaces in fibers Let $S$ be a scheme and let $X$ be a quasi-separated algebraic space over $S$. Does there exist an open subspace of $X$ which is a scheme and which is dense in each fiber $X_s$, $s \in S$?
I am happy to make the following additional assumptions:
(1) $S$ is the spectrum of a complete dvr with algebraically closed residue field and $X$ is flat and of finite type over $S$.
(2) All fibers of $X$ are geometrically irreducible.
(3) The generic fiber of $X$ is geometrically normal and the special fiber of $X$ is geometrically integral (then $X$ is normal).
 A: A counter example to the question in the presence of (1), (2), and (3) is to take $\mathbf{G}_{m, S}$ and divide out by the etale equivalence relation given by $(x, s) \sim (x, s)$ and for $s \not = 0$ also $(x, s) \sim (-x, s)$ (assume the residue characteristic of your dvr is not $2$). Then the resulting morphisms
$$
\mathbf{G}_{m, S} \longrightarrow X \longrightarrow S
$$
have the following properties: the first is surjective \'etale and the second is smooth of relative dimension 1 and both $\mathbf{G}_{m, S}$ and $X$ have geometrically integral fibres over $S$. The schematic locus of X is exactly the complement of the special fibre. It seems to me the "problem" is that the degree of the function field extensions induced between the fibres of $\mathbf{G}_{m, S}$ and $X$ drops from $2$ to $1$ from the generic to the special fibre.
PS: As Jason points out on locally Noetherian separated algebraic spaces the schematic locus contains all codimension $1$ points, see Tag 0ADD
A: There are quasi-separated counterexamples.  If you assume that $X$ is separated, then this is true.  The basic reference is Donald Knutson's book.
MR0302647 (46 #1791) 
Knutson, Donald 
Algebraic spaces. 
Lecture Notes in Mathematics, Vol. 203.  
Springer-Verlag, Berlin-New York, 1971. vi+261 pp. 
If $X$ is separated, then by Chow's Lemma, pp. 192-193, there exists a projective birational morphism $g:\widetilde{X}\to X$ such that the composite morphism $\widetilde{X}\to S$ is a quasi-projective morphism.  Since $X$ is normal, by the valuative criterion of properness and by Zariski's Main Theorem, $g$ is an isomorphism over a dense open subset $U$ of $X$ that contains all codimension one points.  In particular, $U$ contains the generic point of the closed fiber.  Thus, $U$ is dense in every fiber, and $U$ is isomorphic to an open subscheme of the quasi-projective $S$-scheme $\widetilde{X}$.  If you use Raynaud-Gruson, then you can probably say more about $U$.
Edit.  The first example that I wrote, Knutson's beautiful example from pp. 9-10, has nonreduced closed fiber.  The OP asks about the case when the closed fiber is geometrically irreducible.  Here is a slightly different counterexample that is quasi-separated, and even locally separated, such that the closed fiber is geometrically irreducible.
Let $S$ be $\text{Spec}\ \mathbb{C}[[s]]$.  Let $U$ be $\text{Spec}\ \mathbb{C}[[s]][t,t^{-1}]$, i.e., the multiplicative group over $S$.  The fiber product $U\times_S U$ is $\text{Spec}\ \mathbb{C}[[s]][t_1,t_1^{-1},t_2,t_2^{-1}]$.  Define $R \subset U\times_S U$ to be the disjoint union of the closed subscheme $\text{Zero}(t_1-t_2)$, i.e., the diagonal, and the locally closed subscheme $U' = \text{Zero}(t_1+t_2)\cap D(s)$.  This is an étale equivalence relation on $U$.  Thus, by Proposition 1.3(b), p. 93, there is an algebraic space $X$ and an étale, surjective morphism $h:U\to X$ such that $U\times_X U$ equals $R$.  Since $R\to U\times_S U$ is a quasi-compact, locally closed immersion, $X\to S$ is quasi-separated and even locally separated.  The closed fiber $X_0=\text{Zero}(s)$ is just $\text{Spec}\ \mathbb{C}[t,t^{-1}]$, so it is geometrically integral.  
No Zariski open subset of $X$ containing the generic point of the closed fiber is a scheme.  If there were such an open, then there would be an open affine $V$.  Since $X$ is integral, also $V$ is integral.  Thus the ring homomorphism $\mathcal{O}_X(V)\to \mathcal{O}_X(V\cap D(s))$ is injective.  But $D(s)\subset X$ is affine, namely $\text{Spec}\ \mathbb{C}[[s]][s^{-1}][t^2, t^{-2}].$  Every fraction of elements in this ring pulls back on $U$ to a rational function that identifies $t$ and $-t$ on the closed fiber $X_0$.  For an open affine $V$, the restriction map $\mathcal{O}_X(V)\to \mathcal{O}_{X_0}(V\cap X_0)$ is surjective.  Since there are elements in $\mathcal{O}_{X_0}(X_0)$ that do not identify $t$ and $-t$, this is a contradiction.
Second edit. My second example above is precisely the same as Count Dracula's example.  
