For what it's worth, it's sufficient that $\mathcal I $ be comeager, or of Lebesgue measure $> 1/2$.

(And it is not sufficient that $\mathcal I $ be nonmeager, nor that $\mathcal I $ be of Lebesgue measure $\ge 1/2$.)

Let $\lambda$ denote the uniform measure on the powerset of $\mathbb N$ (also known as the Lebesgue measure and the fair-coin measure). Let $\tau$ be the product topology of the discrete topology on $\{0,1\}$.

In order to be able to conclude that $\mathbb N$ can be partitioned into finitely many sets from $\mathcal I$ (in fact, two),

• it is sufficient that $\mathcal I $ be comeager with respect to $\tau$;
• it is sufficient to have $\lambda(\mathcal I)> 1/2$.

These observations are sharp, as the example $\mathcal I:=\{X: 17\in X\}$ shows: it is not sufficient that $\mathcal I $ be nonmeager, nor that $\mathcal I $ be of Lebesgue measure $\ge 1/2$, because it is necessary that $$ \bigcup\mathcal I = \mathbb N. $$