If the function $f: \mathbf{N} \times \mathbf{Z} \rightarrow \mathbf{Z}$ satisfies the following conditions:
(1) $f(0,0)=1, f(0,1)=1$;
(2) For any $k \notin\{0,1\}, f(0, k)=0$;
(3) For each $n \geqslant 1$ and integer $k$, $f(n, k)=f(n-1, k)+f(n-1, k-2 n)$. $(2009$, Turkish Mathematical Olympiad)
Solution 1 Below we prove a lemma using mathematical induction.

Lemma $f(n, k)$ is the coefficient of $x^{k}$ in $(1+x) \prod_{i=1}^{n}\left(1+x^{2 i}\right)$.

Proof When $n=0$, conditions (1) and (2) are obviously satisfied.
Assume that the conclusion holds for $n=m$.
When $n=m+1$, the coefficient of $x^{k}$ in $(1+x) \prod_{i=1}^{m+1}\left(1+x^{2 i}\right)$ is the sum of the coefficients of $x^{k}$ and $x^{k-2 n}$ in $(1+x) \prod_{i=1}^{m}\left(1+x^{2 i}\right)$, which satisfies condition (2).

Therefore, $f(n, k)$ is the coefficient of $x^{k}$ in $(1+x) \prod_{i=1}^{n}\left(1+x^{2 i}\right)$.
Returning to the original problem.
Notice that
$$
\begin{array}{l}
\sum_{k=0}^{n^{2}+n+1} f(n, k) x^{k}=(1+x) \prod_{i=1}^{n}\left(1+x^{2 i}\right) \\
=x^{1+2+4+\cdots+2 n}\left(1+\frac{1}{x}\right) \prod_{i=1}^{n}\left(1+\frac{1}{x^{2 i}}\right) \\
=x^{n^{2}+n+1}\left(1+\frac{1}{x}\right) \prod_{i=1}^{n}\left[1+\left(\frac{1}{x}\right)^{2 i}\right] \\
=x^{n^{2}+n+1} \sum_{k=0}^{n^{2}+n+1} f(n, k) x^{-k} \\
=\sum_{k=0}^{n^{2}+n+1} f(n, k) x^{n^{2}+n+1-k} . \\
\text { Compare }
\end{array}
$$

Comparing the coefficients of $x^{k}$, we have
$$
f(n, k)=f\left(n, n^{2}+n+1-k\right) \text {. }
$$

In the polynomial $\sum_{k=0}^{n^{2}+n+1} f(n, k) x^{k}$, setting $x=1$, we get
$$
\begin{array}{l}
2^{n+1}=\sum_{k=0}^{n^{2}+n+1} f(n, k) \\
=\sum_{k=0}^{c_{k+1}^{2}} f(n, k)+\sum_{k=c_{k}^{2}+1+1}^{n^{2}+n+1} f(n, k) \\
=\sum_{k=1}^{c_{k+1}^{2}} f(n, k)+\sum_{k=0}^{c_{k+1}^{2+1}} f(n, k) \\
=2 \sum_{k=0}^{c_{k+1}^{2}+1} f(n, k) \text { (by equation (1). }
\end{array}
$$

Thus, $\sum_{k=0}^{C_{n+1}^{2}} f(n, k)=2^{n}$.
Therefore, $\sum_{k=0}^{c_{200}} f(2008, k)=2^{2008}$.
Solution 2 (1) Let $S_{n}=\{1,2 \times 1,2 \times 2, \cdots, 2 \times n\}$.
First, we prove by mathematical induction that $f(n, m)$ represents the number of subsets $T \subseteq S_{n}$ such that the sum of elements in $T$ is $m$.
When $n=0$, conditions (1) and (2) are obviously satisfied.
Assume that the conclusion holds for $n=k$.
When $n=k+1$, for each $T \subseteq S_{n}$, we always have
$2(k+1) \in T$ or $2(k+1) \notin T$.
When $2(k+1) \in T$, the number of such $T$ is
$f(k, m-2(k+1))$;
When $2(k+1) \notin T$, the number of such $T$ is $f(k, m)$.
Therefore, the total number of $T$ is
$f(k, m-2(k+1))+f(k, m)=f(k, m)$.
Thus, $\sum_{k=0}^{c_{2} \text { map }} f(2008, k)$ represents the number of subsets $T \subseteq S_{2008}$

such that the sum of elements in $T$ is in $\left[0, \mathrm{C}_{200}^{2}\right]$.
Second, since the sum of all elements in $S_{2008}$ is $2 \mathrm{C}_{2000}^{2} +1$, for any $T$ that satisfies the condition, $S_{2008}-T$ does not satisfy the condition; for any $T$ that does not satisfy the condition, $S_{2008}-T$ satisfies the condition.

Hence, the sets that satisfy the condition and those that do not satisfy the condition are in one-to-one correspondence.

Therefore, $\sum_{k=0}^{c} f(2008, k)$ should be half of the number of subsets of $S_{2008}$, i.e.,
$$
\sum_{k=0}^{c_{2} 2 \infty} f(2008, k)=\frac{2^{2000}}{2}=2^{2008} .
$$