= Solution
Put $\bar i=2n+1-i$ and let $E_{ij}$ denote a <matrix unit>. The antidiagonal matrix satisfies $J^{-1}=J$, so the orthogonal condition is $A=-JA^TJ$, or entrywise $A_{ij}=-A_{\bar j\bar i}$. Consequently the diagonal <Cartan subalgebra> is
$$
\mathfrak t=\{H(t)=\operatorname{diag}(t_1,\ldots,t_n,-t_n,\ldots,-t_1)\},\qquad\varepsilon_i(H(t))=t_i.
$$
We first take $n\geq2$. Since $[H,E_{ab}]=(H_{aa}-H_{bb})E_{ab}$, the <matrix root basis of the even orthogonal Lie algebra> gives
$$
\begin{aligned}\mathfrak g_{\varepsilon_i-\varepsilon_j}&=\mathbb C(E_{ij}-E_{\bar j\bar i})&& (i\ne j),\\\mathfrak g_{\varepsilon_i+\varepsilon_j}&=\mathbb C(E_{i\bar j}-E_{j\bar i})&& (i<j),\\\mathfrak g_{-\varepsilon_i-\varepsilon_j}&=\mathbb C(E_{\bar j i}-E_{\bar i j})&& (i<j).
\end{aligned}
$$
The zero <weight space> is $\mathfrak t$; its <basis> is $H_i=E_{ii}-E_{\bar i\bar i}$. No vector of weight $2\varepsilon_i$ occurs, because the corresponding matrix unit is forced to be its own negative. These spaces account for $n+2n(n-1)=n(2n-1)$ dimensions, exhausting the <Special orthogonal Lie algebra>. Thus its decomposition as a $\mathfrak t$-module and its <Dn root system> are
$$
\boxed{\mathfrak g=\mathfrak t\oplus\bigoplus_{\alpha\in R}\mathfrak g_\alpha,\qquad R=\{\pm\varepsilon_i\pm\varepsilon_j:1\leq i<j\leq n\}.}
$$
Each nonzero <root space> has <dimension> one; the zero weight has multiplicity $n$.
The upper triangular <root spaces> are precisely those of weights $\varepsilon_i-\varepsilon_j$ and $\varepsilon_i+\varepsilon_j$ with $i<j$. Therefore
$$
\boxed{R^+=\{\varepsilon_i-\varepsilon_j,\ \varepsilon_i+\varepsilon_j:i<j\},\qquad\alpha_i=\varepsilon_i-\varepsilon_{i+1}\ (1\leq i<n),\quad\alpha_n=\varepsilon_{n-1}+\varepsilon_n.}
$$
These $n$ roots form the <simple roots>. To check that they generate the chosen positive system with nonnegative coefficients, write $\varepsilon_i-\varepsilon_j=\sum_{k=i}^{j-1}\alpha_k$. For $j<n$,
$$
\varepsilon_i+\varepsilon_j=\sum_{k=i}^{j-1}\alpha_k+2\sum_{k=j}^{n-2}\alpha_k+\alpha_{n-1}+\alpha_n,
$$
and for $j=n$ use $\varepsilon_i+\varepsilon_n=\sum_{k=i}^{n-2}\alpha_k+\alpha_n$. Empty sums are zero. Their <linear independence> and these expansions identify the base of the <root system>.
Equip the real span of the roots in $\mathfrak t^*$ with the <inner product> making the $\varepsilon_i$ orthonormal. All roots then have squared length two. The <coroots> of $\varepsilon_i\pm\varepsilon_j$ are $H_i\pm H_j$, so the <fundamental weights of Dn>, characterized by $\omega_i(\alpha_j^\vee)=\delta_{ij}$, are
$$
\boxed{\begin{aligned}\omega_i&=\varepsilon_1+\cdots+\varepsilon_i&& (1\leq i\leq n-2),\\\omega_{n-1}&=\tfrac12(\varepsilon_1+\cdots+\varepsilon_{n-1}-\varepsilon_n),\\\omega_n&=\tfrac12(\varepsilon_1+\cdots+\varepsilon_{n-1}+\varepsilon_n).
\end{aligned}}
$$
Taking their pairings with $H_j-H_{j+1}$ and $H_{n-1}+H_n$ verifies the defining identities directly. For each pair $i<j$, the sum of its two positive roots is $2\varepsilon_i$. Hence the <half-sum of positive roots> is
$$
\boxed{\rho=\frac12\sum_{\alpha\in R^+}\alpha=\sum_{i=1}^n(n-i)\varepsilon_i=\sum_{i=1}^n\omega_i.}
$$
The <Dynkin diagram> has a chain ending at $\alpha_{n-2}$, which is joined to both $\alpha_{n-1}$ and $\alpha_n$. Every edge is a single bond because the root lengths agree; the nonzero off-diagonal simple-root pairings are minus one. The labelled <Dn Dynkin diagram> and its low-rank versions are shown below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-3-dynkin-diagrams.png]
{title=Labelled D-type Dynkin diagrams, including the low-rank cases}
For $n=2$, $\alpha_1=\varepsilon_1-\varepsilon_2$ and $\alpha_2=\varepsilon_1+\varepsilon_2$ are orthogonal, giving two disconnected <A1 root systems>. For $n=3$, the diagram is the chain $\alpha_2-\alpha_1-\alpha_3$, of type <A3 root system>; the displayed weight formulas still apply. If $n=1$ is allowed, $\mathfrak{so}_2=\mathfrak t$ is one-dimensional and abelian: there are no roots or fundamental weights, and $\rho=0$.
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