Given a finite poset $\mathcal P$, how small can a family $\mathcal F$ of subsets of $[n]$ be such that $\mathcal F$ does not contain an induced copy of $\mathcal P$, but $\mathcal F\cup\{X\}$ contains such a copy for all $X\in\mathcal P([n])\setminus\mathcal F$? This is known as the induced saturation number of $\mathcal P$, denoted by $\text{sat}^*(n,\mathcal P)$. The main conjecture in the area is that the induced saturation number for any poset is either bounded, or linear. In this paper we establish linearity for the induced saturation number of the 4-point poset $\mathcal N$. Previously, it was known that $2\sqrt n\leq\text{sat}^*(n,\mathcal N)\leq 2n$. We show that $\text{sat}^*(n,\mathcal N)\geq\frac{n+6}{4}$. A crucial role in the proof is played by a structural feature of $\mathcal N$-saturated families, namely that if the family contains two antichains, one completely above the other, then it must also contain a `middle' point -- greater than one antichain and less than the other.