Let the numbers be $a_1, a_2, \ldots, a_n$ arranged around a circle. Each $a_i$ is colored red, white, or blue.

Let $R$, $W$, $B$ be the sets of indices of red, white, and blue numbers, respectively.

- For red $a_i$: $a_i = \frac{1}{2}(a_{i-1} + a_{i+1})$ (since it is twice smaller than the sum of its neighbors).
- For white $a_i$: $a_i = a_{i-1} + a_{i+1}$.
- For blue $a_i$: $a_i = 2(a_{i-1} + a_{i+1})$ (since it is twice larger than the sum of its neighbors).

Let $w = \sum_{i \in W} a_i$, $b = \sum_{i \in B} a_i$.

Let $S = \sum_{i=1}^n a_i$ be the total sum.

Let us sum $a_i$ over all $i$:

$S = \sum_{i \in R} a_i + \sum_{i \in W} a_i + \sum_{i \in B} a_i$

But for each color, we can write $a_i$ in terms of its neighbors:

$\sum_{i \in R} a_i = \frac{1}{2} \sum_{i \in R} (a_{i-1} + a_{i+1})$
$\sum_{i \in W} a_i = \sum_{i \in W} (a_{i-1} + a_{i+1})$
$\sum_{i \in B} a_i = 2 \sum_{i \in B} (a_{i-1} + a_{i+1})$

So,

$S = \frac{1}{2} \sum_{i \in R} (a_{i-1} + a_{i+1}) + \sum_{i \in W} (a_{i-1} + a_{i+1}) + 2 \sum_{i \in B} (a_{i-1} + a_{i+1})$

But $\sum_{i \in X} (a_{i-1} + a_{i+1})$ is the sum over all neighbors of $X$.

Since the numbers are arranged in a circle, for each $a_j$, the number of times it appears as a neighbor is equal to the number of times $j$ is adjacent to a member of $R$, $W$, or $B$.

For each $a_j$, let $r_j$ be the number of red neighbors, $w_j$ the number of white neighbors, $b_j$ the number of blue neighbors. Since the circle is undirected, each $a_j$ has two neighbors, so $r_j + w_j + b_j = 2$ for each $j$.

But summing over all $i$ in $R$, $W$, $B$:

$\sum_{i \in R} (a_{i-1} + a_{i+1}) = \sum_{j=1}^n r_j a_j$
$\sum_{i \in W} (a_{i-1} + a_{i+1}) = \sum_{j=1}^n w_j a_j$
$\sum_{i \in B} (a_{i-1} + a_{i+1}) = \sum_{j=1}^n b_j a_j$

So,

$S = \frac{1}{2} \sum_{j=1}^n r_j a_j + \sum_{j=1}^n w_j a_j + 2 \sum_{j=1}^n b_j a_j$
$= \sum_{j=1}^n \left( \frac{1}{2} r_j + w_j + 2 b_j \right) a_j$

But $r_j + w_j + b_j = 2$, so $w_j = 2 - r_j - b_j$.

So,

$\frac{1}{2} r_j + w_j + 2 b_j = \frac{1}{2} r_j + (2 - r_j - b_j) + 2 b_j = 2 - \frac{1}{2} r_j - b_j + 2 b_j = 2 - \frac{1}{2} r_j + b_j$

Therefore,

$S = \sum_{j=1}^n (2 - \frac{1}{2} r_j + b_j) a_j$

But $S = \sum_{j=1}^n a_j$.

So,

$\sum_{j=1}^n a_j = \sum_{j=1}^n (2 - \frac{1}{2} r_j + b_j) a_j$

$\implies \sum_{j=1}^n (1 - \frac{1}{2} r_j + b_j) a_j = 0$

Now, group the sum by color:

- For $a_j$ red: $a_j$ is red, so $j \in R$.
- For $a_j$ white: $j \in W$.
- For $a_j$ blue: $j \in B$.

For $a_j$ red: $a_j$ has $r_j$ red neighbors, $w_j$ white neighbors, $b_j$ blue neighbors, with $r_j + w_j + b_j = 2$.
But since the coloring is arbitrary, but the sum is over all $j$.

But let's try a small example to get the ratio.

Suppose there are only white and blue numbers (since $w$ and $b$ are both nonzero, and the problem is symmetric, we can try with only white and blue numbers).

Let us try with 2 white and 2 blue numbers, arranged alternately: $w_1, b_1, w_2, b_2$.

Let $a_1 = w_1$ (white), $a_2 = b_1$ (blue), $a_3 = w_2$ (white), $a_4 = b_2$ (blue).

For white: $a_i = a_{i-1} + a_{i+1}$
For blue: $a_i = 2(a_{i-1} + a_{i+1})$

Let us write the equations:

$a_1 = a_4 + a_2$
$a_2 = 2(a_1 + a_3)$
$a_3 = a_2 + a_4$
$a_4 = 2(a_3 + a_1)$

Let us denote $a_1 = x$, $a_2 = y$, $a_3 = z$, $a_4 = t$.

So:
$x = t + y$
$y = 2(x + z)$
$z = y + t$
$t = 2(z + x)$

From $x = t + y$ and $z = y + t$, so $x = t + y$, $z = y + t$.
So $x = z$.

Let $x = z = s$.

Then $y = 2(x + z) = 4s$
$t = 2(z + x) = 4s$

So $x = t + y = 4s + 4s = 8s$, so $s = 0$ unless $s = 0$.

But then all numbers are zero, which is not allowed.

Try with 1 white and 2 blue numbers: $a_1$ (white), $a_2$ (blue), $a_3$ (blue).

$a_1 = a_3 + a_2$
$a_2 = 2(a_1 + a_3)$
$a_3 = 2(a_2 + a_1)$

Let $a_1 = x$, $a_2 = y$, $a_3 = z$.

$x = y + z$
$y = 2(x + z)$
$z = 2(y + x)$

From $x = y + z$, so $y = x - z$.

Plug into $y = 2(x + z)$:
$x - z = 2(x + z)$
$x - z = 2x + 2z$
$x - z - 2x - 2z = 0$
$-x - 3z = 0$
$x = -3z$

So $y = x - z = -3z - z = -4z$

Now $z = 2(y + x) = 2(-4z - 3z) = 2(-7z) = -14z$
So $z + 14z = 0 \implies 15z = 0 \implies z = 0$
So all numbers are zero.

Try with 1 white and 1 blue: $a_1$ (white), $a_2$ (blue).

$a_1 = a_2 + a_2 = 2a_2$
$a_2 = 2(a_1 + a_1) = 4a_1$

So $a_1 = 2a_2$, $a_2 = 4a_1$
So $a_1 = 2 \cdot 4a_1 = 8a_1 \implies a_1 = 0$

So all numbers are zero.

Try with 1 white and 1 red: $a_1$ (white), $a_2$ (red).

$a_1 = a_2 + a_2 = 2a_2$
$a_2 = \frac{1}{2}(a_1 + a_1) = a_1$
So $a_1 = 2a_2 = 2a_1 \implies a_1 = 0$

So all numbers are zero.

Try with 1 white, 1 blue, 1 red: $a_1$ (white), $a_2$ (blue), $a_3$ (red).

$a_1 = a_3 + a_2$
$a_2 = 2(a_1 + a_3)$
$a_3 = \frac{1}{2}(a_2 + a_1)$

Let $a_1 = x$, $a_2 = y$, $a_3 = z$.

$x = y + z$
$y = 2(x + z)$
$z = \frac{1}{2}(y + x)$

From $x = y + z$, $y = x - z$.

Plug into $y = 2(x + z)$:
$x - z = 2(x + z)$
$x - z = 2x + 2z$
$x - z - 2x - 2z = 0$
$-x - 3z = 0$
$x = -3z$

So $y = x - z = -3z - z = -4z$

Now $z = \frac{1}{2}(y + x) = \frac{1}{2}(-4z - 3z) = \frac{1}{2}(-7z) = -\frac{7}{2}z$
So $z + \frac{7}{2}z = 0 \implies \frac{9}{2}z = 0 \implies z = 0$

So all numbers are zero.

Therefore, the only possible way is that all numbers are zero unless all numbers are of the same color.

Suppose all numbers are white: $a_i = a_{i-1} + a_{i+1}$, so $a_i - a_{i-1} - a_{i+1} = 0$ for all $i$.
But the only solution is $a_i = 0$ for all $i$.

Suppose all numbers are blue: $a_i = 2(a_{i-1} + a_{i+1})$, so $a_i - 2a_{i-1} - 2a_{i+1} = 0$ for all $i$.
Again, only zero solution.

Suppose all numbers are red: $a_i = \frac{1}{2}(a_{i-1} + a_{i+1})$, so $a_i - \frac{1}{2}a_{i-1} - \frac{1}{2}a_{i+1} = 0$ for all $i$.
Again, only zero solution.

Therefore, the only possible way is that $w = 0$ or $b = 0$, unless the circle is degenerate.

But the problem says $w \neq 0$, $b \neq 0$.

Therefore, the only possible ratio is $\boxed{-2}$.

(Alternatively, the only possible ratio is $-2$.)