I thought I could figure this out just be reading up on root systems and weights, but I couldn't find an easy way to explain it. I'll ask a specific version of the question, but this observation might hold more generally.
Let $\Phi$ be the irreducible root system of type $D_n$, say for $n > 3$, with positive roots $\Phi^+$ and simple roots $\Delta$. Index the simple roots according to their Dynkin diagram as follows: $$ \begin{aligned} &\>2 \\[-1ex] &\,\,\mid \\[-1ex] 1\,\!-\!&\;3\!-\!4\!-\dotsb-\!n \end{aligned} $$ Think of $\Phi^+$ as a subset of $\mathbf{N}^n$ with $\Delta$ as your set standard basis vectors—so the vector corresponding to a root is it's dimension vector. If you want to count the root vectors in $\Phi^+$ that have the form $\langle 0,0,0,\ast,\dotsc,\ast\rangle$, you'll find that there are $\frac{1}{2}(n-2)(n-3)$ of them. You can see this from the Dynkin diagram because it's the same as counting positive roots of a type $A_{n-3}$ root system. Now if you count the number of root vectors of the form $\langle 1,1,2,\ast,\dotsc,\ast\rangle$ you'll find there are also $\frac{1}{2}(n-2)(n-3)$ of them. Why does this happen?
There is similar symmetry counting the other root vectors too. There are as many root vectors of the form $\langle 0,1,0,\ast,\dotsc,\ast\rangle$ as there are of the form $\langle 1,1,1,\ast,\dotsc,\ast\rangle$, and there are as many root vectors of the form $\langle 0,0,1,\ast,\dotsc,\ast\rangle$ as of the form $\langle 1,0,0,\ast,\dotsc,\ast\rangle$, but these two are a bit more obvious from the diagram.
I'm used to thinking of these roots vectors as indecomposable quiver representations, so if there's an explanation from that perspective, that'd help me out fine, but maybe it won't be as helpful to future students who find this question.