$2 \cdot 69$ For each integer $n \geqslant 2$, determine the minimum value of the sum $a_{0}+a_{1}+\cdots+a_{n}$ when $a_{0}, a_{1}, \cdots, a_{n}$ are non-negative numbers, and
$$
a_{0}=1, a_{i} \leqslant a_{i+1}+a_{i+2}, i=0,1, \cdots, n-2
$$
［Solution］First, consider a special case: $a_{0}=1$, and $a_{i}=a_{i+1}+a_{i+2}, i=0,1, \cdots, n-2$.
In this case, we denote $u=a_{n-1}, v=a_{n}$, then
$$
a_{k}=F_{n-k} u+F_{n-k-1} v, k=0,1, \cdots, n-1 .
$$

where $F_{0}=0, F_{1}=1, F_{i+2}=F_{i}+F_{i+1}$, i.e., $F_{i}$ is the $i$-th Fibonacci number, and the sum
$$
a_{0}+a_{1}+\cdots+a_{n}=\left(F_{n+2}-1\right) u+F_{n+1} v .
$$

Since
$$
1=a_{0}=F_{n} u+F_{n-1} v,
$$

and it is easy to verify that
$$
\frac{F_{n+2}-1}{F_{n}} \leqslant \frac{F_{n+1}}{F_{n-1}},
$$

the sum reaches its minimum value $\frac{F_{n+2}-1}{F_{n}}$ when $v=0, u=\frac{1}{F_{n}}$.
Let $M_{n}=\frac{F_{n+2}-1}{F_{n}}$.
We will now prove that $M_{n}$ is the desired minimum value.
Clearly, for each $n \geqslant 2$, the sum
$$
a_{0}+a_{1}+a_{2}+\cdots+a_{n} \geqslant a_{0}+a_{1}+a_{2} \geqslant 2 a_{0}=2 .
$$

Since $M_{2}=\frac{F_{4}-1}{F_{2}}=2, M_{3}=\frac{F_{5}-1}{F_{3}}=2$, when $n=2$, $M_{2}$ is the minimum value; when $n=3$, $M_{3}$ is the minimum value.

Now fix an integer $n \geqslant 4$. Assume that for each $k, 2 \leqslant k \leqslant n-1$, if the non-negative sequence $c_{0}, c_{1}, \cdots, c_{k}$ satisfies the condition $c_{0}=1, c_{i} \leqslant c_{i+1}+c_{i+2}, i=0,1, \cdots, k-2$, then the sum $c_{0}+c_{1}+\cdots+c_{k} \geqslant M_{k}$.
Consider the non-negative sequence $a_{0}, a_{1}, \cdots, a_{n}$ that satisfies the given conditions.
If $a_{1}, a_{2}>0$, the sum $a_{0}+a_{1}+\cdots+a_{n}$ can be expressed in the following two forms:
$$
\begin{aligned}
a_{0}+a_{1}+\cdots+a_{n} & =1+a_{1}\left(1+\frac{a_{2}}{a_{1}}+\cdots+\frac{a_{n}}{a_{1}}\right) \\
& =1+a_{1}+a_{2}\left(1+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n}}{a_{2}}\right) .
\end{aligned}
$$

The sums inside the parentheses satisfy the induction hypothesis conditions $(k=n-1, k=n-2)$, so we have
$$
\begin{array}{l}
a_{0}+a_{1}+\cdots+a_{n} \geqslant 1+a_{1} M_{n-1}, \\
a_{0}+a_{1}+\cdots+a_{n} \geqslant 1+a_{1}+a_{2} M_{n-2} .
\end{array}
$$

Obviously, (1) and (2) also hold when $a_{1}=0$ or $a_{2}=0$.
By $a_{2} \geqslant 1-a_{1}$, (2) can be transformed into
$$
a_{0}+a_{1}+\cdots+a_{n} \geqslant 1+M_{n-2}+a_{1}\left(1-M_{n-2}\right) \text {, }
$$

Let $\quad f(x)=1+M_{n-1} x$,
$$
g(x)=\left(1+M_{n-2}\right)+\left(1-M_{n-2}\right) x,
$$

From (1) and (3), we get
$$
a_{0}+a_{1}+\cdots+a_{n} \geqslant \max \left\{f\left(a_{1}\right), g\left(a_{1}\right)\right\},
$$

Since $f$ is increasing and $g$ is decreasing, their graphs intersect at a unique point $(\widetilde{x}, \widetilde{y})$, and $\max \{f(x), g(x)\} \geqslant \widetilde{y}$ for each real number $x$.

From (4) and (5), we get
$$
a_{0}+a_{1}+\cdots+a_{n} \geqslant \widetilde{y},
$$

On the other hand, when $x=\frac{F_{n-1}}{F_{n}}$, it is easy to verify that
$$
f(x)=g(x)=\frac{F_{n+2}-1}{F_{n}}=M_{n} .
$$

Therefore
$$
\widetilde{y}=M_{n} .
$$

From (6), we get
$$
a_{0}+a_{1}+\cdots+a_{n} \geqslant M_{n} .
$$

In summary, for each $n \geqslant 2$, the minimum value of the sum $a_{0}+a_{1}+\cdots+a_{n}$ is $M_{n}$.