26. For each integer $n \geqslant 2$, determine the minimum value of 
$$
a_{0}+a_{1}+\cdots+a_{n}
$$
satisfying the conditions
$$
a_{0}=1, a_{i} \leqslant a_{i+1}+a_{i+2}, i=0,1, \cdots, n-2
$$
where $a_{0}, a_{1}, \cdots, a_{n}$ are non-negative numbers.
Consider the special case where $a_{i}=a_{i+1}+a_{i+2}$.
Let $u=a_{n-1}, v=a_{n}$, then we have $a_{k}=F_{n-k} u + F_{n-k-1} v, k=0,1, \cdots, n-1$. Here, $F_{i}$ is the $i$-th Fibonacci number $\left(F_{0}=0, F_{1}=1, F_{i+2}=F_{i}+F_{i+1}\right)$. The sum is
$$
a_{0}+a_{1}+\cdots+a_{n}=\left(F_{n+2}-1\right) u+F_{n+1} v.
$$

Given $1=a_{0}=F_{n} u+F_{n-1} v$, it is easy to verify that $\frac{F_{n+2}-1}{F_{n}} \leqslant \frac{F_{n+1}}{F_{n-1}}$.
To minimize the sum, let $v=0, u=\frac{1}{F_{n}}$, then the sequence
$$
a_{k}=\frac{F_{n-k}}{F_{n}}, k=0,1, \cdots, n
$$

has the sum $M_{n}=\frac{F_{n+2}-1}{F_{n}}$.
Thus, we conjecture that $M_{n}$ is the required minimum value. We prove this by mathematical induction:

For each $n$, the sum $a_{0}+a_{1}+\cdots+a_{n}$ is at least 2, because $a_{0}=1, a_{0} \leqslant a_{1}+a_{2}$.

When $n=2, n=3$, the value of (2) is 2. So, the conjecture holds in these two cases.

Now fix an integer $n \geqslant 4$, and assume that for each $k, 2 \leqslant k \leqslant n-1$, the non-negative sequence $c_{0}, c_{1}, \cdots, c_{k}$ satisfying the conditions $c_{0}=1, c_{i} \leqslant c_{i+1}+c_{i+2}, i=0,1, \cdots, k-2$ has the sum $c_{0}+c_{1}+\cdots+c_{k} \geqslant M_{k}$.

Consider the non-negative sequence $a_{0}, a_{1}, \cdots, a_{n}$ satisfying the given conditions. If $a_{1}, a_{2}>0$ and $a_{1}+\cdots+a_{n}$ can be expressed in the following two forms:
$$
\begin{array}{l}
a_{0}+a_{1}+\cdots+a_{n}=1+a_{1}\left(1+\frac{a_{2}}{a_{1}}+\cdots+\frac{a_{n}}{a_{1}}\right) \\
=1+a_{1}+a_{2}\left(1+\frac{a_{3}}{a_{2}}+\cdots+\frac{a_{n}}{a_{2}}\right).
\end{array}
$$

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

If $a_{1}=0$ or $a_{2}=0$, (3) and (4) also hold. Since $a_{2} \geqslant 1-a_{1}$, from (4) we get
$$
a_{0}+a_{1}+\cdots+a_{n} \geqslant 1+M_{1-}+a_{1}\left(1-M_{n-2}\right).
$$

Combining (3) and the above inequality, we get
$$
a_{0}+a_{1}+\cdots+a_{2} \geqslant \operatorname{max}\left\{f\left(a_{1}\right), g\left(a_{1}\right)\right\}.
$$

Here, $f, g$ are linear functions $f(x)=1+M_{n-1} x, g(x)=\left(1+M_{n-2}\right)+\left(1-M_{n-2}\right) x$. Since $f$ is increasing and $g$ is decreasing, their graphs intersect at a unique point $(\tilde{x}, \tilde{y})$, and
$\max \{f(x), g(x)\} \geqslant \tilde{y}$ for each real number $x$.
Let $x=\frac{F_{n-1}}{F_{n}}$, by (1) we know that $x$ is the value of $a_{1}$. It is easy to prove
$$
f\left(\frac{F_{n-1}}{F_{n}}\right)=g\left(\frac{F_{n-1}}{F_{n}}\right)=\frac{F_{n+2}-1}{F_{n}}=M_{n}.
$$

Therefore, $\tilde{y}=M_{n}$. By (5) and (6), we have $a_{0}+\cdots+a_{n} \geqslant \tilde{y}=M_{n}$. Thus, for each $n \geqslant 2, M_{n}$ is the minimum value of the sum $a_{0}+a_{1}+\cdots+a_{n}$.

Note: The author of this problem also provided a more sophisticated second proof. The approach is as follows: If $\left(a_{0}, a_{1}, \cdots, a_{n}\right)$ is a sequence satisfying the given conditions $(n \geqslant 4)$, and there exists at least one strict inequality $a_{i}<a_{i+1}+a_{i+2}$, then a slight modification of the sequence produces a new sequence with a smaller sum that also satisfies the given conditions. Therefore, any optimal sequence satisfies $a_{i}=a_{i+1}+a_{i+2}$ for all $i$. Hence, it can only be the sequence defined in (1).