Let $I_{n}=\left(\frac{a_{n}}{n}, \frac{a_{n}+1}{n}\right), n=1,2, \cdots, 2002$. If there exists a real number $x \in \bigcap_{n=1}^{2002} I_{n}$, then the proposition is proved.

For this, let $L=\max _{1 \leqslant n \leqslant 2002} \frac{a_{n}}{n}, U=\min _{1 \leqslant n \leqslant 2002} \frac{a_{n}+1}{n}$. We want to prove $L < U$. To do this, we need to show that for any $n, m \in \{1, 2, \cdots, 2002\}$, the following inequality holds:
$$m a_{n} < n (a_{m} + 1). \tag{1}$$

We will prove this by induction on $m$.

Base Case: For $m = 1$, the inequality becomes $a_{n} < n (a_{1} + 1)$. Since $a_{1} \geq 0$, this is clearly true.

Inductive Step: Assume that for some $m \geq 1$, the inequality holds for all $n \in \{1, 2, \cdots, 2002\}$. We need to show that the inequality holds for $m+1$.

- If $m > n$, then by the induction hypothesis, we have $(m-n) a_{n} < n (a_{m-n} + 1)$. By the given condition, $n (a_{m-n} + a_{n}) \leq n a_{m}$. Therefore, $m a_{n} < n (a_{m} + 1)$, and inequality (1) holds.
- If $m < n$, by the induction hypothesis, we have $m a_{n-m} < (n - m) (a_{m} + 1)$. By the given condition, $m a_{n} \leq m (a_{m} + a_{n-m} + 1)$. Adding these two inequalities, we get $m a_{n} < n (a_{m} + 1)$, and inequality (1) also holds.

In conclusion, for $n, m \in \{1, 2, \cdots, 2002\}$, inequality (1) always holds.