For a chemical reaction,
$$\Delta G = -R T \ln K_{eq}$$
$$\implies K_{eq} = e^{\frac{-\Delta G}{RT}} $$
Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$
For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the temperature and pressure at which a chemical reaction is taking place is coincidentally just that needed to achieve $\Delta G = 0$.
On the other hand, in finding phase transition temperatures (melting and boiling points), we are sweeping through a range of temperatures specifically to find the points at which the phases coexist, i.e., where their free energies are the same and thus where $\Delta G = 0$. Thus these are equilibria for which $K_{eq}=1$.
*We can never exactly get $\Delta G = 0$, because we can never exactly adjust T and p; consequently, we can never exactly have $K_{eq}=1$. Thus what I mean (and presumably what your textbook means) when it says "$K_{eq}=1$" is that it is so close to $1$ that it's within the size of the fluctuations about equilibrium.
**I say "for some chemical reactions" because in some cases the reactants or products might break down before the required temperature is reached. E.g., suppose you were trying to find a reaction involving glucose at which $K_{eq} =1$, and did a theoretical extrapolation in which you found that this occurred at, say, 10,000 K. Well, glucose doesn't exist at that temperature.
Note #1: $\Delta G = -R T \ln K_{eq}$ applies to processes carred out at constant T and p. If the process were instead at, say, constant T and V, then the equilibrium condition would be $\Delta A = -R T \ln K_{eq}$. [And don't be confused: Constant T and p doesn't mean a specific T and p. T and p can be whatever you please, so long as the reactants and products can exist at those conditions. Rather, constant T and p just means that, whatever T and p are, they stay fixed during the reaction.]
Note #2 (on time to attain equilibrium): The above has nothing to do with whether or not you can get to equilbrium, which is addressed in the question linked in Ivan's comment on your post. That question pertains to any equilibrium, regardless of the value of the equilibrium constant (whether it's 1 or 10.303 or 4.8 x 10^-22, or any other value). Instead, the above is about whether or not the value of the equilibrium constant can be 1.
Note #3 (on enantiomers in an achiral environment): Andrew raises an interesting point that the reaction:
$$\ce{A_L <=> A_D},$$
where $\ce{A_L}$ and $\ce{A_D}$ are two different enantiomers of compound A, will have $\Delta G = 0$, and thus $K_{eq}=1$, in an achiral environment because, in an achiral environment, the two compounds are chemically identical. And as they are chemically identical, this will obtain at all temperatures and pressures at which the compound exists. I.e., in an achiral environment, chemically,
$$\ce{A_L <=> A_D}$$
reduces to:
$$\ce{A <=> A}$$.
But: While enantiomers are chemically identical, the weak nuclear force that exists within their nuclei can recognize handedness, and thus the restriction in the first paragraph still applies: $K_{eq}$ won't exactly equal 1, but can be immeasurably (at least with current technology) close to 1. [I haven't done the calculation to determine if the difference in energy caused by the weak nuclear force puts the deviation from 1 within the, say, the 90% fluctation range about equilibrium for a representative pair of enantiomers at room temperature.]
