Sure, here is the translation:

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II. (50 points) Let $f(x)$ be a function defined on the set of natural numbers $\mathbf{N}$ and taking values in $\mathbf{N}$, satisfying: for any two distinct natural numbers $a$ and $b$,
$$
f(a) + f(b) - f(a + b) = 2009.
$$
(1) Find $f(0)$;
(2) Let $a_{1}, a_{2}, \cdots, a_{100}$ be 100 distinct natural numbers, find
$$
\begin{array}{l}
f\left(a_{1}\right) + f\left(a_{2}\right) + \cdots + f\left(a_{100}\right) - \\
f\left(a_{1} + a_{2} + \cdots + a_{100}\right);
\end{array}
$$
(3) Does there exist a function $f(x)$ that satisfies the given conditions and such that $f(2009) = 2009^2$? Prove your conclusion.

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(1) Take $a=0, b=1$, then $a \neq b$.
By the condition, we have
$$
f(0)+f(1)-f(0+1)=2.009 \text {. }
$$

So, $f(0)=2009$.
(2) Proof: For any natural number $n$ greater than 1, and $n$ distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{n}\right)-f\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
=2009 n-2009 .
\end{array}
$$

We use induction on $n$.
When $n=2$, the conclusion is obviously true.
Assume the conclusion holds for $n=k(k \geqslant 2)$.
Then, for $n=k+1$, let the $k+1$ distinct natural numbers be $a_{1}, a_{2}, \cdots, a_{k+1}$, where $a_{1}<a_{2}<\cdots<a_{k+1}$. By the induction hypothesis, we have
$$
\begin{array}{c}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{k+1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+ \\
f\left(a_{1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) .
\end{array}
$$

Since $0 \leqslant a_{1}<a_{2}<\cdots<a_{k+1}$, we have
$$
a_{1}<a_{2}+a_{3}+\cdots+a_{k+1} \text {. }
$$

Thus, $a_{1}$ and $a_{2}+a_{3}+\cdots+a_{k+1}$ are distinct.
$$
\begin{array}{l}
\text { Then } f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+f\left(a_{1}\right) \\
=f\left(a_{1}+a_{2}+a_{3}+\cdots+a_{k+1}\right)+2009 \text {. } \\
\text { Hence } f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{k+1}\right)- \\
f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+ \\
f\left(a_{1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+2009 \\
=2009(k+1)-2009 \text {. } \\
\end{array}
$$

Therefore, the conclusion holds for $n=k+1$.
So, the conclusion is true.
Taking $n=100$, we get
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{10}\right)-f\left(a_{1}+a_{2}+\cdots+a_{10}\right) \\
=200900-2009=198891 . \\
\text { (3) Let } f(x)=2008 x+2009 .
\end{array}
$$

We now prove that the function $f(x)$ defined this way satisfies the condition.
Indeed, for any two distinct natural numbers $a$ and $b$, we have
$$
\begin{array}{l} 
f(a)+f(b)-f(a+b) \\
=(2008 a+2009)+(2008 b+2009)- \\
(2008 a+2008 b+2009) \\
= 2008 a+2008 b-(a+b)+2009 \\
= 2009, \\
\text { Hence } f(2009)=2008 \times 2009+2009 \\
= 2009^{2} .
\end{array}
$$