Letting $n=2018$, we will show that, if every region has at least one non-yellow vertex, then every circle contains at most $n+\lfloor\sqrt{n-2}\rfloor-2$ yellow points. In the case at hand, the latter equals $2018+44-2=2060$, contradicting the hypothesis.

Consider the natural geometric graph $G$ associated with the configuration of $n$ circles. Fix any circle $C$ in the configuration, let $k$ be the number of yellow points on $C$, and find a suitable lower bound for the total number of yellow vertices of $G$ in terms of $k$ and $n$. It turns out that $k$ is even, and $G$ has at least
$$\begin{equation*} k+2\binom{k / 2}{2}+2\binom{n-k / 2-1}{2}=\frac{k^{2}}{2}-(n-2) k+(n-2)(n-1) \tag{*} \end{equation*}$$
yellow vertices. The proof hinges on the two lemmata below.

Lemma 1. Let two circles in the configuration cross at $x$ and $y$. Then $x$ and $y$ are either both yellow or both non-yellow.

Proof. This is because the numbers of interior vertices on the four arcs $x$ and $y$ determine on the two circles have like parities.
In particular, each circle in the configuration contains an even number of yellow vertices.

Lemma 2. If $\text{x y}$, $\text{y z}$, and $\text{z x}$ are circular arcs of three pairwise distinct circles in the configuration, then the number of yellow vertices in the set $\{x, y, z\}$ is odd.

Proof. Let $C_{1}, C_{2}, C_{3}$ be the three circles under consideration. Assume, without loss of generality, that $C_{2}$ and $C_{3}$ cross at $x$, $C_{3}$ and $C_{1}$ cross at $y$, and $C_{1}$ and $C_{2}$ cross at $z$. Let $k_{1}$, $k_{2}, k_{3}$ be the numbers of interior vertices on the three circular arcs under consideration. Since each circle in the configuration, different from the $C_{i}$, crosses the cycle $\text{x y y z z x}$ at an even number of points (recall that no three circles are concurrent), and self-crossings are counted twice, the sum $k_{1}+k_{2}+k_{3}$ is even.

Let $Z_{1}$ be the colour $z$ gets from $C_{1}$ and define the other colours similarly. By the preceding, the number of bichromatic pairs in the list $\left(Z_{1}, Y_{1}\right),\left(X_{2}, Z_{2}\right),\left(Y_{3}, X_{3}\right)$ is odd. Since the total number of colour changes in a cycle $Z_{1}-Y_{1}-Y_{3}-X_{3}-X_{2}-Z_{2}-Z_{1}$ is even, the number of bichromatic pairs in the list $\left(X_{2}, X_{3}\right),\left(Y_{1}, Y_{3}\right),\left(Z_{1}, Z_{2}\right)$ is odd, and the lemma follows.

We are now in a position to prove that (*) bounds the total number of yellow vertices from below. Refer to Lemma 1 to infer that the $k$ yellow vertices on $C$ pair off to form the pairs of points where $C$ is crossed by $k / 2$ circles in the configuration. By Lemma 2, these circles cross pairwise to account for another $2\binom{k / 2}{2}$ yellow vertices. Finally, the remaining $n-k / 2-1$ circles in the configuration cross $C$ at non-yellow vertices, by Lemma 1, and Lemma 2 applies again to show that these circles cross pairwise to account for yet another $2\binom{n-k / 2-1}{2}$ yellow vertices. Consequently, there are at least (*) yellow vertices.

Next, notice that $G$ is a plane graph on $n(n-1)$ degree 4 vertices, having exactly $2 n(n-1)$ edges and exactly $n(n-1)+2$ faces (regions), the outer face inclusive (by Euler's formula for planar graphs).

Lemma 3. Each face of $G$ has equally many red and blue vertices. In particular, each face has an even number of non-yellow vertices.

Proof. Trace the boundary of a face once in circular order, and consider the colours each vertex is assigned in the colouring of the two circles that cross at that vertex, to infer that colours of non-yellow vertices alternate.

Consequently, if each region has at least one non-yellow vertex, then it has at least two such. Since each vertex of $G$ has degree 4, consideration of vertex-face incidences shows that $G$ has at least $n(n-1) / 2+1$ non-yellow vertices, and hence at most $n(n-1) / 2-1$ yellow vertices. (In fact, Lemma 3 shows that there are at least $n(n-1) / 4+1 / 2$ red, respectively blue, vertices.)

Finally, recall the lower bound (*) for the total number of yellow vertices in $G$, to write $n(n-1) / 2-1 \geqslant k^{2} / 2-(n-2) k+(n-2)(n-1)$, and conclude that $k \leqslant n+\lfloor\sqrt{n-2}\rfloor-2$, as claimed in the first paragraph.

The first two lemmata in Solution 1 show that the circles in the configuration split into two classes: Consider any circle $C$ along with all circles that cross $C$ at yellow points to form one class; the remaining circles then form the other class. Lemma 2 shows that any pair of circles in the same class cross at yellow points; otherwise, they cross at non-yellow points.

Call the circles from the two classes white and black, respectively. Call a region yellow if its vertices are all yellow. Let $w$ and $b$ be the numbers of white and black circles, respectively; clearly, $w+b=n$. Assume that $w \geqslant b$, and that there is no yellow region. Clearly, $b \geqslant 1$, otherwise each region is yellow. The white circles subdivide the plane into $w(w-1)+2$ larger regions - call them white. The white regions (or rather their boundaries) subdivide each black circle into black arcs. Since there are no yellow regions, each white region contains at least one black arc.

Consider any white region; let it contain $t \geqslant 1$ black arcs. We claim that the number of points at which these $t$ $\operatorname{arcs}$ cross does not exceed t-1. To prove this, consider a multigraph whose vertices are these black arcs, two vertices being joined by an edge for each point at which the corresponding arcs cross. If this graph had more than t-1$edges, it would contain a cycle, since it has$t$ vertices; this cycle would correspond to a closed contour formed by black sub-arcs, lying inside the region under consideration. This contour would, in turn, define at least one yellow region, which is impossible.

Let $t_{i}$ be the number of black arcs inside the $i^{\text{th}}$ white region. The total number of black arcs is $\sum_{i} t_{i}=2 w b$, and they cross at $2\binom{b}{2}=b(b-1)$ points. By the preceding,
$$b(b-1) \leqslant \sum_{i=1}^{w^{2}-w+2}\left(t_{i}-1\right)=\sum_{i=1}^{w^{2}-w+2} t_{i}-\left(w^{2}-w+2\right)=2 w b-\left(w^{2}-w+2\right)$$
or, equivalently, $(w-b)^{2} \leqslant w+b-2=n-2$, which is the case if and only if $w-b \leqslant\lfloor\sqrt{n-2}\rfloor$. Consequently, $b \leqslant w \leqslant(n+\lfloor\sqrt{n-2}\rfloor) / 2$, so there are at most $2(w-1) \leqslant n+\lfloor\sqrt{n-2}\rfloor-2$ yellow vertices on each circle - a contradiction.