Three, (50 points) Find the number of positive integers $t$ not exceeding 2009, such that for all natural numbers $n$, $\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}$ is coprime with 2009.

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$$
\begin{array}{l}
\text { Three, let } \\
\begin{array}{l} 
x= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}=\frac{1}{\sqrt{t}} \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}(\sqrt{t})^{2 k+1}, \\
\begin{array}{l} 
y= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k} t^{k}=\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k}(\sqrt{t})^{2 k} . \\
\text { Then } \sqrt{t} x+y=\sum_{k=0}^{2 n+1} \mathrm{C}_{2 n+1}^{k}(\sqrt{t})^{k} \\
=(\sqrt{t}+1)^{2 n+1}, \\
\sqrt{t} x-y=-\sum_{k=1}^{2 n+1} \mathrm{C}_{2 n+1}^{k}(-\sqrt{t})^{k} \\
=(\sqrt{t}-1)^{2 n+1} . \\
\text { (1) }+(2)
\end{array} \\
x=\frac{1}{2 \sqrt{t}}\left[(\sqrt{t}+1)^{2 n+1}+(\sqrt{t}-1)^{2 n+1}\right]=x_{n} .
\end{array}
\end{array}
$$

Therefore, $x_{0}=1, x_{1}=3+t$,
$$
\begin{aligned}
x_{n+2}= & {\left[(\sqrt{t}+1)^{2}+(\sqrt{t}-1)^{2}\right] x_{n+1}-} \\
& (\sqrt{t}+1)^{2}(\sqrt{t}-1)^{2} x_{n},
\end{aligned}
$$

which simplifies to $x_{n+2}=2(t+1) x_{n+1}-(t-1)^{2} x_{n}$.
Multiplying (1) and (2) gives
$$
t x^{2}-y^{2}=(t-1)^{2 n+1} \text {. }
$$

Given $2009=7^{2} \times 41$, we have
$$
(x, 2009)=1 \Leftrightarrow 7 \nmid x, 41 \nmid x \text {. }
$$
(1) $\times$ (2) yields
$$
t x^{2}-y^{2}=(t-1)^{2 n+1} \text {. }
$$

(1) $7 \nmid x$. We will discuss this under modulo 7.

If $t \equiv 0$, then $x_{n} \equiv \mathrm{C}_{2 n+1}^{1}=2 n+1$, and $x_{3} \equiv 0$. This is a contradiction.

If $t \equiv 1$, then $x_{n} \equiv \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}=2^{2 n} \not\equiv 0$, which satisfies the condition.
If $t \equiv 4$, then $x_{1} \equiv 0$, which is a contradiction.
From equation (3), if $t \equiv 2,3,5,6$, then $\{x_{n}\}$ modulo 7 is a periodic sequence, and for all positive integers $n, x_{n} \neq 0$, which satisfies the condition.
Thus, the required $t \equiv 1,2,3,5,6 \pmod{7}$.
(2) $41 \nmid x$. We will discuss this under modulo 41.

If $t \equiv 0$, then $x_{n} \equiv \mathrm{C}_{2 n+1}^{1}=2 n+1$, and $x_{21} \equiv 0$, which is a contradiction.

If $t \equiv 1$, then $x_{n} \equiv \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}=2^{2 n} \not\equiv 0$, which satisfies the condition.

If $t \equiv -1$, then $x_{0}=1, x_{1} \equiv 2, x_{n+2} \equiv -4 x_{n}$, which satisfies the condition.

If $t \equiv 10, -20$, then $x_{2}=t^{2}+10 t+5 \equiv t^{2}+10 t-200=(t-10)(t+20) \equiv 0$, which is a contradiction.
$$
\begin{array}{l} 
\text { If } t \equiv \pm 3, \pm 17, \pm 19, 6, 11, -7, -15, \\
x_{11}=\sum_{k=0}^{10} \mathrm{C}_{2 k}^{2 k+1} t^{k} \\
\equiv t^{10}+5 t^{9}-t^{8}-20 t^{7}+7 t^{6}-7 t^{5}+t^{4}+ \\
4 t^{3}+13 t^{2}+18 t-20 \\
\equiv\left(t^{2}-3^{2}\right)\left(t^{2}-17^{2}\right)\left(t^{2}-19^{2}\right)(t-6) . \\
(t-11)(t+7)(t+15) \\
\equiv 0,
\end{array}
$$

which is a contradiction.
When $t-1=s \equiv \pm 3, \pm 6, \pm 7, \pm 11, \pm 12, \pm 13, \pm 14, \pm 15, \pm 17, \pm 19$, considering the remainders of $y^{2}$ and $s^{2 n+1}$ modulo 41, we have
$$
y^{2}+(t-1)^{2 n+1}=y^{2}+s^{2 n+1} \not\equiv 0 .
$$

From equation (1), $x \neq 0$.
Thus, $t \equiv 4,7,8,12,13,14,15,16,18,20, -1, -2, -4, -5, -6, -8, -9, -10, -11, -12, -13, -14, -16, -18$ satisfy the condition.

From equation (3), if $t \equiv 2,5,9, -4, -8, -9$, then $\{x_{n}\}$ modulo 41 is a periodic sequence, and for all positive integers $n, x_{n} \neq 0$, which satisfies the condition.
Thus, the required
$$
t \equiv 1,2,4,5,7,8,9,12,13,14,15,16,18,
$$
$20, -1, -2, -4, -5, -6, -8, -9, -10, -11, -12, -13, -14, -16, -18 \pmod{41}$.

From (1), (2), and the Chinese Remainder Theorem, the number of $t$ that satisfy the condition is $\frac{5 \times 28}{7 \times 41} \times 2009=980$.
$$