1. **Rewrite the polynomial $P(x)$:**
$$P(x) = 3\sum_{k=0}^{9}x^k + 2\sum_{k=10}^{1209}x^k + \sum_{k=1210}^{146409}x^k$$
This can be expressed as:
$$P(x) = 3(1 + x + x^2 + \cdots + x^9) + 2(x^{10} + x^{11} + \cdots + x^{1209}) + (x^{1210} + x^{1211} + \cdots + x^{146409})$$

2. **Consider the problem in $\mathbb{Z}_{11}$:**
We need to find the smallest $n$ such that $x^n - 1$ is divisible by $(x^{16} + 1)P(x)$ in $\mathbb{Z}_{11}[x]$.

3. **Check divisibility by $x - 1$:**
Since $x^n - 1$ must be divisible by $x - 1$, we check if $x = 1$ is a root of $P(x)$ and $x^{16} + 1$:
$$P(1) = 3 \cdot 10 + 2 \cdot 1200 + 134200 = 3 \cdot 10 + 2 \cdot 1200 + 134200 = 30 + 2400 + 134200 = 136630 \equiv 0 \pmod{11}$$
$$x^{16} + 1 = 1^{16} + 1 = 2 \not\equiv 0 \pmod{11}$$
Thus, $x^{16} + 1$ is not divisible by $x - 1$.

4. **Construct $Q(x)$:**
Multiply $x - 1$ by $P(x)$ and adjust the constant term:
$$Q(x) = 4x^{146410} + 4x^{1210} + 4x^{10} - 1$$
Let $y = 4x^{10}$. By Fermat's Little Theorem, $4^{11} \equiv 4 \pmod{11}$:
$$Q(y) = y^{14641} + y^{121} + y - 1$$

5. **Find the smallest $m$ such that $y^m - 1$ is divisible by $Q(y)$:**
We need $y^m - 1 = h(y)Q(y)$. Using the recursion $c_i = c_{i-1} + c_{i-121} + c_{i-14641}$, we find:
$$c_i = \binom{d_1 + d_2 + d_3}{d_1, d_2, d_3}$$
The smallest $i$ such that $c_i = 1$ and $c_j = 0$ for $i > j > (i-14641)$ is $i = 121^3 - 1$.

6. **Convert back to $x$:**
$$y = 4x^{10} \implies x^{10(121^3 - 1)} - 1$$
This polynomial is divisible by $Q(x)$ but not by $x^{16} + 1$. The next polynomial is:
$$x^{20(121^3 - 1)} - 1$$

7. Verify the exponent:
$$20(121^3 - 1) = 20 \times 1771560 = 35431200$$

The final answer is $\boxed{35431200}$