1. Claim: The answer is no.

2. Reasoning: Since $P(x)$ is an odd polynomial with nonnegative coefficients, it must be of the form:
$$P(x) = a_{2m+1}x^{2m+1} + a_{2m-1}x^{2m-1} + \cdots + a_1x$$
where $a_i \geq 0$ for all $i$.

3. Derivative: The derivative of $P(x)$ is:
$$Q(x) = P'(x) = (2m+1)a_{2m+1}x^{2m} + (2m-1)a_{2m-1}x^{2m-2} + \cdots + a_1$$

4. Key Claim: Suppose $(a, P(a))$ and $(b, P(b))$ are consecutive points in the sequence. Then $a$ and $b$ must have different signs.

5. Proof: The tangent line at $(a, P(a))$ has the equation:
$$y = Q(a)(x - a) + P(a)$$
Substituting $x = b$, we get:
$$P(b) = Q(a)(b - a) + P(a)$$
Rearranging, we have:
$$Q(a) = \frac{P(b) - P(a)}{b - a}$$
Since $P(x)$ is an odd function, we can write:
$$P(b) = a_{2m+1}b^{2m+1} + a_{2m-1}b^{2m-1} + \cdots + a_1b$$
and similarly for $P(a)$. Thus:
$$Q(a) = a_{2m+1} \frac{b^{2m+1} - a^{2m+1}}{b - a} + a_{2m-1} \frac{b^{2m-1} - a^{2m-1}}{b - a} + \cdots + a_1$$

6. Grouping Terms: Grouping the terms of degree $2m$ together, we get:
$$(2m+1)a^{2m} - b^{2m} - b^{2m-1}a - \cdots - a^{2m}$$
Suppose $a$ and $b$ have the same sign. If $|a| > |b|$, then the expression in the brackets is positive, and if $|a| < |b|$, then the expression is negative.

7. Consistent Sign: Similarly, if we group the terms of degrees $2m-2, 2m-4, \cdots, 0$ together, we obtain a consistent sign for all the expressions, which means that the equation:
$$Q(a) - \frac{P(b) - P(a)}{b - a} = 0$$
has no solutions, leading to a contradiction.

8. Conclusion: Since $n$ is odd, this means that $A_n$ and $A_1$ have to lie on the same side of the $y$-axis, which is a contradiction.

$\blacksquare$