4. A magician and his assistant have a deck of cards, all of which have the same back, and the front is one of 2017 colors (each color has 1000000 cards). The magic trick is: the magician first leaves the room, the audience arranges $n$ face-up cards in a row on the table, the magician's assistant then flips $n-1$ of these cards over in their original positions, leaving only one card face-up, and then the magician enters the room, observes the cards on the table, and selects a face-down card, guessing its color. Find the minimum value of $n$, such that the magician and his assistant can complete this trick according to some predetermined strategy.
4. When $n=2018$, the magician and the assistant can agree that for $i=1,2, \cdots, 2017$, if the assistant retains the $i$-th card face up, the magician will guess that the 2018-th card is of the $i$-th color. This way, the magic trick can be completed.

Assume for some positive integer $n \leqslant 2017$, the magic trick can be completed.

For convenience, use $(a, i)$ to denote that the $a$-th card is of the $i$-th color, where $1 \leqslant a \leqslant n, 1 \leqslant i \leqslant 2017$.

For $n$ cards, note that the number of cards of each color is 1000000 (greater than $n$), so there are $2017^{n}$ different states (each position has 2017 possible colors). Each state is equivalent to a set of the form
$$
\left\{\left(1, i_{1}\right),\left(2, i_{2}\right), \cdots,\left(n, i_{n}\right)\right\}
$$
where $i_{1}, i_{2}, \cdots, i_{n} \in\{1,2, \cdots, 2017\}$.
For any $a, i$, since when the magician sees $(a, i)$, they can always guess at least one $(b, j)(b \neq a)$, and the number of states containing $(a, i)$ and $(b, j)$ is only $2017^{n-2}$, the total number of different states the magician can guess (i.e., $2017^{n}$) does not exceed $2017 \times 2017^{n-2} n$.
Thus, $n \geqslant 2017$.
This indicates that $n$ can only be 2017. Therefore, from any $(a, i)$, the $(b, j)(b \neq a)$ that can be guessed is unique.

Therefore, draw an edge from $(a, i)$ to $(b, j)$, denoted as $(a, i) \rightarrow(b, j)$, and call $a, b$ "associated" (association is mutual, i.e., $b, a$ are also associated), and any two different guesses $(a, i) \rightarrow(b, j)$ and $(c, k) \rightarrow(d, l)$
cannot correspond to the same state.
This indicates that at least one of $c, d$ belongs to $\{a, b\}$.
The following lemma is proved.
Lemma If $(a, i) \rightarrow(b, j)$, then there must exist $k \neq i$ such that $(b, j) \rightarrow(a, k)$.
Proof Suppose $(b, j) \rightarrow(c, k)$. If $c \neq a$ or $(c, k)=(a, i)$, then a state containing $(a, i), (b, j), (c, k)$ has two guesses $(a, i) \rightarrow(b, j)$ and $(b, j) \rightarrow(c, k)$, which is a contradiction.
Thus, there must exist a $k \neq i$ such that
$(b, j) \rightarrow(a, k)$.
The lemma is proved.
Assume 1 and 2 are associated.
Note that 3 must be associated with another number $a$. By the previous discussion, $a \in\{1,2\}$, assume $a=1$.
At this point, for any $b(4 \leqslant b \leqslant 2017)$, since $b$ must be associated with another number $c$, and 1 and 2 are associated, 1 and 3 are associated, thus $c \in\{1,2\}, c \in\{1,3\}$. Therefore, $c=1$, i.e., 1 and $b$ are associated.
This indicates that 2, 3, ..., 2017 are all associated with 1.
Thus, for any $t \in\{2,3, \cdots, 2017\}$, combining the lemma, there exist $i_{t}, j_{t} \in\{1,2, \cdots, 2017\}$ such that $\left(1, i_{t}\right) \rightarrow\left(t, j_{t}\right)$.
Using the lemma repeatedly, there exist $k_{t}, l_{t} \in\{1,2, \cdots, 2017\}\left(k_{t} \neq i_{t}, l_{t} \neq j_{t}\right)$ such that
$\left(t, j_{t}\right) \rightarrow\left(1, k_{t}\right) \rightarrow\left(t, l_{t}\right)$.
Thus, $i_{2}, k_{2}, i_{3}, k_{3}, \cdots, i_{2017}, k_{2017}$ are all different, but $2 \times 2016 > 2017$, which is a contradiction. This indicates that the case $n=2017$ cannot complete the magic trick.
In summary, the minimum value of $n$ is 2018.