The answer is YES. We will prove by induction that the procedure can be applied for any positive integer $n$.

The statement is true for $n=1$. Suppose that it is also true for $n=k-1 \geq 1$, which means there exists a way to change every number not exceeding $k-1$ from black to white, which we call process $A$.

We shall prove the statement for $k$ with some cases as follows:

1. If $k$ is not a square-free number. Put $k=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{v}^{a_{v}}$ and $k' = p_{1} p_{2} \ldots p_{v}$ with $p_{1}, p_{2}, \ldots, p_{v}$ primes. Note that the number of times the color of $k$ and $k'$ are changed are equal (since the set of integers not coprime with them are the same), then $k'$ can change the color if and only if $k$ can. So after making process $A$ with the sequence $1,2, \ldots, k-1$, the number $k$ will change color from black to white.

2. If $k$ is a square-free number. In case after making process $A$, number $k$ changes color, then we are done. Otherwise, put $k=p_{1} p_{2} \ldots p_{t}$ and we consider process $B$ to select all divisors other than $1$ of $k$. After that, number $k$ will be affected $2^{t}-1$ times and will change the color.

Denote $k'$ as some divisor greater than $1$ of $k$, and $k'$ has $s<t$ prime divisors. Take some number $c$ which is a multiple of $k'$ but coprime to $k / k'$. After process $A$, number $c$ has color white. And after process $B$, the number of times the color of $c$ is changed is $\left(2^{t}-1\right)-\left(2^{t-s}-1\right)=2^{t}-2^{t-s}$ which is an even number. This implies that $c$ will not change color.

So in all cases, we always can find a process to change the color of all numbers from $1$ to $k$ from black to white.

Finally, by replacing $n=10^{6}$, the problem is solved. $\square$