14. For each integer $n \geqslant 2$, determine the maximum value of the sum $a_{0}+a_{1}+a_{2}+\cdots+a_{n}$ satisfying the conditions $a_{0}=1, a_{i} \geqslant a_{i+1}+a_{i+2}, i=0$. $1, \cdots, n-2$ (where $a_{0}, a_{1}, \cdots a_{n}$ are non-negative).
14. If there is 1 place where the inequality holds: $a_{i}>a_{i+1}+a_{i+2}$, then let $a_{i+1}^{\prime}=a_{i}-a_{i+2}, a_{j}^{\prime}=a_{1}(j \neq i)$.

Then $a_{i}^{\prime}=a_{i+1}^{\prime}+a_{i+2}^{\prime}, a_{i+1}^{\prime}>a_{i+1} \geqslant a_{t+2}^{\prime}+a_{i+3}^{\prime}$, so $a_{0}^{\prime}, a_{1}^{\prime}, \cdots, a_{n}^{\prime}$ still satisfy the problem's conditions while $a_{0}^{\prime}+a_{1}^{\prime}+\cdots+a_{n}^{\prime}>a_{0}+a_{1}+\cdots+a_{n}$.
Therefore, when the sum is maximized, $a_{i}=a_{i+1}+a_{i+2}$ holds for any $0 \leqslant i \leqslant n-2$.
At this point, using the recursive relation, we can prove that $a_{k}=F_{n-k} \cdot a_{n-1}+F_{n-k-1} \cdot a_{n}$, where $F_{2}$ is the $i$-th Fibonacci number $\left(F_{0}=0, F_{1}=1, F_{i+2}=F_{2+1}+F_{i}\right)$, then $a_{0}+a_{1}+\cdots+a_{n}=$ $\left(F_{n+2}-1\right) a_{n-1}+F_{n-1} a_{n}$.
Also, $a_{0}=1=F_{n} \cdot a_{n-1}+F_{n-1} \cdot a_{n}$, so $a_{n}=\frac{1-F_{n} \cdot a_{n-1}}{F_{n} \cdot 1}$.
Thus, $a_{0}+a_{1}+\cdots+a_{n}=\frac{F_{n+2} \cdot F_{n-1}-F_{n+1} \cdot F_{n}-F_{n-1}}{F_{n-1}} \cdot a_{n-1}+\frac{F_{n+1}}{F_{n-1}}$
$$
\begin{array}{l}
=\frac{(-1)^{n}-F_{n-1}}{F_{n-1}} \cdot a_{n-1}+\frac{F_{n+1}}{F_{n-1}} \\
\leqslant \frac{(-1)^{n}-F_{n-1}}{F_{n-1}} \cdot 0+\frac{F_{n+1}}{F_{n-1}}=\frac{F_{n+1}}{F_{n-1}} .
\end{array}
$$

Thus, the maximum value is $\frac{F_{n+1}}{F_{n-1}}$, which is achieved when $a_{n-1}=0, a_{n}=\frac{1}{F_{n-1}}, a_{i}=a_{i+1}+a_{i+2}(0 \leqslant i \leqslant n-2)$.