[Proof] We prove a stronger conclusion: among integers satisfying $1 \leqslant a_{0}<a_{1}<\cdots<a_{n} \leqslant 2 n$ -3, there must exist $i, j, k, l$, all not equal to $n$, such that
$$a_{i}+a_{j}=a_{k}+a_{l}=a_{n}.$$

We use proof by contradiction. Assume that there do not exist $i, j, k, l$ satisfying the above equation, and let
$$A=\left\{a_{0}, a_{1}, \cdots, a_{n-1}\right\}.$$

When $a_{n}=2 p+1$ (an odd number), consider $p$ pairs of numbers $(1+i, 2 p-i), i=0,1, 2, \cdots, p-1$. Since the sum of each pair is $2 p+1=a_{n}$, the number of pairs where both numbers belong to $A$ is at most one. Since these $p$ pairs of $2 p$ numbers include all integers from 1 to $2 p$, the elements of set $A$ must all appear in these $p$ pairs. Therefore, the number of elements in set $A$ is
$$n \leqslant p+1,$$
$$2 n \leqslant 2 p+2=a_{n}+1 \leqslant(2 n-3)+1,$$ which is a contradiction.
When $a_{n}=2 p$ (an even number), consider $p-1$ pairs of numbers $(1+i, 2 p-1-i)$, $i=0,1, \cdots, p-2$. Since the sum of each pair is $2 p=a_{n}$, the number of pairs where both numbers belong to $A$ is at most one. Since these $p-1$ pairs of $2 p-2$ numbers include all integers from 1 to $2 p-1$ except $p$, all elements of set $A$ except possibly $p$ must appear in these $p-1$ pairs. Therefore, the number of elements in set $A$ is
$$\begin{array}{l} n \leqslant p+1, \\ 2 n \leqslant 2 p+2=a_{n}+2 \leqslant(2 n-3)+2, \text { which is a contradiction. } \end{array}$$

Thus, the original proposition is proved.