Does every reductive group scheme admit a maximal torus? A theorem of Grothendieck states that any smooth reductive algebraic group over a field $k$ admits a maximal torus over $k$. My question concerns what happens for schemes.

Let $S$ be a scheme and let $G$ be a smooth reductive group scheme over $S$. Does $G$ admit a maximal torus over $S$?

Given that this is Grothendieck we are talking about, I imagine if he knew how to prove the result over schemes he would have done so, in particular I'm willing to believe that the answer is "no". However I don't know any explicit counter-examples.
I have a specific application in mind where my scheme $S$ is not too badly behaved, for example I can assume that $S$ is Noetherian, affine, regular with $\mathrm{Pic}(S)=0.$ If anyone knows any positive results in such special cases, I would also be very interested.
 A: I might well be missing something here, but:
Consider $S = \mathbb{P}^2$ and $E$ the tangent bundle to $\mathbb{P}^2$. Set $G = GL(E)$. If $T$ is a maximal torus of $G$ then,for every point $x \in \mathbb{P}^2$, we have a maximal torus $T_x$. The eigenspaces of $T_x$ form two points in $\mathbb{P}(E)$; let $\Lambda \subset \mathbb{P}(E)$ be the set of eigenspaces of the maximal torii. Then $\Lambda \to \mathbb{P}^2$ is a double cover. Since $\mathbb{P}^2$ is simply connected, $\Lambda$ has two connected components. Let $L_1$ and $L_2$ be the sub-line bundles of $E$ spanned by these components. Then $E = L_1 \oplus L_2$. But a standard computation with Chern classes shows that $E$ is not the direct sum of two line bundles.
Of course, this example is neither affine nor has vanishing Pic, as you last paragraph requests.

I now have an regular affine example with vanishing Pic. Take $X = \{ (a,b,c,x,y,z) : ax+by+cz=1 \} \subset \mathbb{C}^6$. I first note that $X$ is simply connected: Projection onto the $(x,y,z)$ plane reveals $X$ to be a rank $2$ affine bundle over $\mathbb{C}^3 \setminus \{ 0 \}$, so $X$ is homotopy equivalent to $\mathbb{C}^3 \setminus 0$, or to $S^5$. 
In this answer, Steve Lansberg shows that $\mathcal{O}(X)$ is a UFD, so all line bundles on $X$ are trivial, but that $X$ possesses a nontrivial vector bundle $E$ of rank $2$. As in the previous answer, if $GL(E)$ had a maximal torus, the eigenspaces of that torus would give a double cover of $X$. Since $X$ is simply connected, that double cover is trivial and the torus is split. Then $E = L_1 \oplus L_2$. Since $L_1$ and $L_2$ are line bundles on $X$, they are trivial, but then $E$ is trivial, a contradiction.
A: I doubt that this is true even in the simplest case $G=GL(E)$, where $E$ over $S$ is a nondecomposable vector bundle.
A: Counterexamples exist even over $S = \mathrm{Spec} \mathbb{Z}$. See, for instance, Lemma 1.1 and Example 6.2 of http://math.stanford.edu/~conrad/papers/redgpZ.pdf or other places of that paper. If $\mathscr{G}$ in Example 6.2 there had a maximal torus over $\mathbb{Z}$, then that maximal torus would be split, contradicting the anisotropy of $G_{\mathbb{R}}$.
