Here is the context in which I heard it. If $G$ is a reductive group over a non-archimedean local field $F$, then let $K$ be the hyperspecial maximal compact subgroup $G(\mathcal{O})$. For example, if $G=SO(n)$, and assuming that $G$ is split, then there is a natural choice of $SO(n)(\mathcal{O})$. What exactly does the last sentence mean? (I know that a hyperspecial maximal compact subgroup exists if and only if $G$ is quasisplit and splits over an unramified extension -- this is stated in Tits' Corvallis article.) The only context I am aware of is in the theory of Lie groups.



What is a "split $SO(n)$"?
In fact, to properly understand what is $SO(n)(\mathcal{O})$, you need to know what are split groups over a ring. A summary of this story goes as follows:


Given a ring $R$, and given a finite Dynkin diagram (<https://en.wikipedia.org/wiki/Dynkin_diagram>), you have a unique simply connected (resp. adjoint) absolutely simple split algebraic group over $R$. For example, for the $A\_n$ diagram, the corresponding algebraic group is $SL\_{n+1}$ (resp. $PGL\_{n+1}\simeq SL\_{n+1}/\mu\_{n+1}$ ) considered as an algebraic group over $R$.


But you also have some groups "in between" the simply connected one and the adjoint one. In the $A\_n$ case, this corresponds to the fact that you could quotient by proper subgroups of $\mu\_{n+1}$.


Finally, your group $K$ in the $A\_n$ case would just be $SL\_{n+1}(\mathcal{O})$, or if working with the adjoint guy, it would be $PGL\_{n+1}(\mathcal{O}) = GL\_{n+1}(\mathcal{O})/\mathcal{O}^{\times} $. Warning here : computing rational points of quotients of algebraic groups is a complicated story. But it's ok for this particular quotient using that $\mathcal{O}$ is a PID. 


Now, $SO(2n+1)(\mathcal{O})$ (resp. $SO(2n)(\mathcal{O})$) is just the group of rational points of a well-chosen quotient of THE simply connected absolutely simple split algebraic group over $\mathcal{O}$ of type $B\_n$ (resp. $D\_n$).


To make this more concrete and link that back to the comments, let us assume that $n$ is odd, or that $2$ is invertible in $\mathcal{O}$. Then actually $SO(n)(\mathcal{O})\simeq \lbrace g\in SL\_n(\mathcal{O})~\vert~ q(g.x) = q(x)\rbrace $, where $q$ is the quadratic form $(x\_{-k},...,x\_k)\mapsto \sum x\_{-i}x\_i$ (we can have $2k+1$ or $2k$ variables, depending on whether or not we wish to include the index $0$).


If you want to know what $SO(n)$ is when $n$ is even and $2$ is not invertible in the base ring, you can for example look in "The Book of Involutions" by Knus, Merkurjev, Rost and Tignol (their definition of the split special orthogonal group works over an arbitrary ring). There is also the Appendix C of "Reductive Group Schemes" by Conrad, for that matter. 


