Suppose $a_i, b_i, c_i$ for $i = 1, 2, \ldots, n$ are $3n$ real numbers in the interval $[0, 1]$. Define the sets
$$S = \{ (i, j, k) \mid a_i + b_j + c_k < 1 \}$$
and
$$T = \{ (i, j, k) \mid a_i + b_j + c_k > 2 \}.$$
We are given that $|S| \geq 2018$ and $|T| \geq 2018$. We aim to find the minimal possible value of $n$.

To establish a lower bound for $n$, consider the projections of the sets $S$ and $T$ onto the coordinate planes. Note that $S_{xy} \cap T_{xy} = \emptyset$, meaning that no pair $(a_i, b_j)$ can simultaneously satisfy $a_i + b_j + c_k < 1$ and $a_i + b_j + c_k > 2$ for any $c_k$.

Thus, we have the inequalities:
$$|S_{xy}| + |T_{xy}| \leq n^2, \quad |S_{yz}| + |T_{yz}| \leq n^2, \quad |S_{zx}| + |T_{zx}| \leq n^2.$$

Applying the Projection Inequality and Hölder's Inequality, we obtain:
$$2 \cdot 2018^{2/3} \leq |S|^{2/3} + |T|^{2/3} \leq |S_{xy}|^{1/3} \cdot |S_{yz}|^{1/3} \cdot |S_{zx}|^{1/3} + |T_{xy}|^{1/3} \cdot |T_{yz}|^{1/3} \cdot |T_{zx}|^{1/3} \leq (|S_{xy}| + |T_{xy}|)^{1/3} (|S_{yz}| + |T_{yz}|)^{1/3} (|S_{zx}| + |T_{zx}|)^{1/3} \leq n^2.$$

Solving for $n$, we get:
$$2 \cdot 2018^{2/3} \leq n^2 \implies n \geq \sqrt{2} \cdot 2018^{1/3} \approx 17.8.$$

Thus, the minimal possible value of $n$ is:
$$n \geq 18.$$

The answer is: \boxed{18}.