Let $a_{1}=\frac{1}{4}, a_{n}=\frac{1}{4}\left(1+a_{n-1}\right)^{2}, n \geqslant 2$. Find the smallest
real number $\lambda$, such that for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2} 002$, we have
$$
\sum_{k=1}^{2002} A_{k} \leqslant \lambda a_{2002} \text {. }
$$

where $A_{k}=\frac{x_{k}-k}{\left(x_{k}+\cdots+x_{2002}+\frac{k(k-1)}{2}+1\right)^{2}}, k \geqslant 1$.
Let $\delta(k)=\frac{1}{2} k(k-1)$. First, we prove a few lemmas.

Lemma 1 For any real numbers $a \geqslant 0, c>0, b>0$, the function $f(x)=\frac{a}{x+b}+\frac{x-c}{(x+b)^{2}}$.
When $x=\frac{(1-a) b+2 c}{1+a}$, it attains the maximum value $\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c}$.
Proof: Let $y=\frac{1}{x+b}$, then
$f(x)=-(b+c) y^{2}+(1+a) y$
$=-(b+c)\left(y-\frac{1}{2} \cdot \frac{1+a}{b+c}\right)^{2}+\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c}$.
Thus, when $y=\frac{1}{2} \cdot \frac{1+a}{b+c}$, i.e., $x=\frac{(1-a) b+2 c}{1+a}$,
$f(x)_{\max }=\frac{1}{4} \cdot \frac{(1+a)^{2}}{b+c}$.

Lemma 2 Let $a_{1}=\frac{1}{4}, a_{n}=\frac{1}{4}\left(1+a_{n-1}\right)^{2}, n \geqslant 2$. Then $a_{n}$ satisfies $0<a_{n}<1$.

Lemma 3 For any $n \geqslant 1, \sum_{k=1}^{n} A_{k} \leqslant \frac{1}{\delta(n+1)+1} a_{n}$, and equality can be achieved.
Proof: By Lemma 1, we have
$$
\begin{array}{l}
\frac{x_{1}-1}{\left(x_{1}+\cdots+x_{n}+1\right)^{2}} \leqslant \frac{1}{4} \cdot \frac{1}{x_{2}+\cdots+x_{n}+2} \\
=\frac{a_{1}}{x_{2}+\cdots+x_{n}+2},
\end{array}
$$

and when $x_{1}=x_{2}+\cdots+x_{n}+3$, it attains the maximum value
$$
\begin{array}{l}
\frac{a_{1}}{x_{2}+\cdots+x_{n}+2} \cdot \\
\frac{a_{1}}{x_{2}+\cdots+x_{n}+2}+\frac{x_{2}-2}{\left(x_{2}+\cdots+x_{n}+2\right)^{2}} \\
\leqslant \frac{1}{4} \cdot \frac{\left(1+a_{1}\right)^{2}}{x_{3}+\cdots+x_{n}+4}=\frac{a_{2}}{x_{3}+\cdots+x_{n}+4},
\end{array}
$$

and when $x_{2}=\frac{\left(1-a_{1}\right)\left(x_{3}+\cdots+x_{n}+4\right)+4}{1+a_{1}}$, it attains the maximum value
$$
\frac{a_{2}}{x_{3}+\cdots+x_{n}+4}.
$$
......
$$
\begin{array}{l}
\frac{a_{n-2}}{x_{n-1}+x_{n}+\delta(n-1)+1}+\frac{x_{n-1}-(n-1)}{\left(x_{n-1}+x_{n}+\delta(n-1)+1\right)^{2}} \\
\leqslant \frac{1}{4} \cdot \frac{\left(1+a_{n-2}\right)^{2}}{x_{n}+\delta(n)+1}=\frac{a_{n-1}}{x_{n}+\delta(n)+1},
\end{array}
$$

and when $x_{n-1}=\frac{\left[\left(1-a_{n-2}\right)\left(x_{n}+\delta(n-1)+1\right)+2(n-1)\right]}{1+a_{n-2}}$
it attains the maximum value $\frac{a_{n-1}}{x_{n}+\delta(n)+1}$.
$$
\begin{array}{l}
\frac{a_{n-1}}{x_{n}+\delta(n)+1}+\frac{x_{n}-n}{\left(x_{n}+\delta(n)+1\right)^{2}} \\
\leqslant \frac{1}{4} \cdot \frac{\left(1+a_{n-1}\right)^{2}}{\delta(n+1)+1}=\frac{a_{n}}{\delta(n+1)+1},
\end{array}
$$

and when $x_{n}=\frac{\left(1-a_{n-1}\right)(\delta(n)+1)+2 n}{1+a_{n-1}}$, it attains the maximum value
$$
\frac{a_{n}}{\delta(n+1)+1}.
$$

By adding (1), (2), ..., (n), we get
$$
\sum_{k=1}^{n} A_{k} \leqslant \frac{1}{\delta(n+1)+1} a_{n},
$$

and when $x_{n}=\frac{\left(1-a_{n-1}\right)(\delta(n)+1)+2 n}{1+a_{n-1}}$,
$$
\begin{array}{l}
x_{n-1}=\frac{\left(1-a_{n-2}\right)\left(x_{n}+\delta(n-1)+1\right)+2(n-1)}{1+a_{n-2}}, \\
\cdots \cdots, \\
x_{2}=\frac{\left(1-a_{1}\right)\left(x_{3}+\cdots+x_{n}+4\right)+4}{1+a_{1}}, \\
x_{1}=x_{2}+\cdots+x_{n}+3
\end{array}
$$

equality holds.
By Lemma 3, we obtain
$$
\lambda=\frac{1}{\delta(2003)+1}=\frac{1}{2003 \times 1001+1}.
$$