A number in combinatorial game theory is a game $x=\{x^L\mid x^R\}$ such that all its options are numbers and there are no $x^L,x^R$ such that $x^L\geq x^R$.
It turns out, after some work, that if $x$ is a number $x^L<x<x^R$ for all $x^L,x^R$. However I can't seem to find (or derive myself) any information about the converse, namely "if $x$ is such that $x^L<x<x^R$ for all $x^L,x^R$, then $x$ is a number".
If a counterexample $x$ is to be found, it must be true by the Simplicity Theorem (ONAG) that no (surreal) number $n$ satisfies $x^L\ngeq n\ngeq x^R$ for all $x^L,x^R$. By using this observation and Theorem 56 in ONAG, we get that if $x$ is a counterexample it must be comparable to all numbers. Unfortunately this is not absurd in and on itself, for example the game $\uparrow :=\{0\mid *\}$ is not a number but is comparable to all numbers (it's greater than $0$ but smaller than all positive numbers). Notice however that $\uparrow$ is not a counterexample, since $0\nless *$.