We may rephrase the condition saying: $\mathcal{N}$ includes the class $\Sigma_0$ of all $E\in \Sigma$, such that $E$ is a null set for some non-atomic probability measure $\mu_E$ on $\Sigma$, or equivalently, of all $E\in \Sigma$ such that there is a non-atomic probability measure supported on $\Omega\setminus E$. This is a condition only on the set $\Omega\setminus E$ as a measurable space with the $\sigma$-algebra induced by $\Sigma$. Re-naming this measure subspace, the question is:
Which measurable space $(\Omega, > \Sigma)$ can support a non-atomic probability measure $\mu$?
There is a number of papers on this matter, that you can easily find on google, and should check, though mainly addressed to spaces with additional topological or metrc sturctures. However, in the general setting of a measurable spaces $(\Omega,\Sigma)$ I think one can actually give a caracterization for the existence of a non-atomic probability measure in terms of the partial order structure of $\Sigma$ (I'm quite sure somebody here has the right reference handy).
A first necessary condition on the structure of $\Sigma$ as a partially ordered set is given by Sierpiński's theorem : a non-atomic probability measure space $(\Omega,\Sigma,\mu)$ is divisible; precisely: $\Sigma$ contains a monotone family $\{E _ \lambda\} _ {\lambda\in[0,1]}$ such that $\mu(E_\lambda)=\lambda$ (in other words, the measure $\mu$, as a function $\mu:\Sigma\to[0,1]$, admits a monotone section $[0,1]\ni\lambda\mapsto E_\lambda\in\Sigma$). And, of course, every set of positive measure also have this property.
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Summarizing: Denoting by $\mathcal{M^*}$ the set of non-atomic probability measures on $(\Omega,\Sigma)$, and $$\Sigma_0:=\{A\in\Sigma: \exists \mu\in\mathcal{M^*} \, s.t. \mu(E)=0\}\, ,$$
a necessary condition for $E_0\in\Sigma_0$ is that it can be embedded into a family $\{E _ \lambda\} _ {\lambda\in[0,1]}$ of subsets of $\Sigma$ strictly increasing by inclusion. In other words, $E_0$ is a minimum element of a chain of $\Sigma$ that is order-isomorphic to $[0,1]$.