Let $a_{1},a_{2},\cdots,a_{N}$ be nonnegative reals, not all $0$. Prove that there exists a sequence $$1=n_{0}<n_{1}<\cdots<n_{k}=N+1$$ of integers such that $$n_{1}a_{n_{0}}+n_{2}a_{n_{1}}+\cdots+n_{k}a_{n_{k-1}}<3(a_{1}+a_{2}+\cdots+a_{N})$$
I need to understand the problem here.
and I fell it is Strange to use probability to can solve it. Can find other methods?
This is basically the same argument as the probability approach used in the original solution, but somewhat simplified and hopefully easier to understand.
To help avoid special cases for the $n_i$, let's start by appending zeroes to the list $a_1,\ldots,a_N,0,\ldots,0$ so it gets length $2^k-1\ge N$. We will make a solution to this consisting of $1=n_0<\cdots<n_{k-1}<n_k=2^k$. To solve the original problem, depending on whether $n_{k-1}>N$ or not, replace $n_k$ with $n_k=N+1$, or remove $n_k$ and replace $n_{k-1}$ with $n_{k-1}=N+1$.
We consider all combinations of $n_1\in\{2,3\}$, $n_2\in\{4,5,6,7\}$, etc.: i.e., $n_i\in\{2^i,\ldots,2^{i+1}-1\}$ for $i=1,\ldots,k-1$. This gives us lots of options for the sequence $n_i$.
Next, we average over all combinations of $n_i$. The average value of $n_i$ is $(3\cdot2^i-1)/2<3\cdot2^{i-1}$ while the average value of $a_{n_i}$ is $(a_{2^i}+\cdots a_{2^{i+1}-1})/2^i$. Since $n_i$ and $n_{i+1}$ are chosen independently, this makes the average value of the term $a_{n_i}n_{i+1}$ $$ \text{E}[a_{n_i}n_{i+1}] =\text{E}[a_{n_i}]\cdot\text{E}[n_{i+1}]<3\cdot(a_{2^i}+\cdots+a_{2^{i+1}-1}) $$ which also holds for the term $i=k-1$ as $n_k=2^k<3\cdot 2^{k-1}$.
Adding all the $a_{n_i}n_{i+1}$ terms together then yields $$ \text{E}[a_{n_0}n_1+\cdots+a_{n_{k-1}}n_k] <3\cdot(a_1+\cdots+a_{2^{k}-1}). $$ Since the bound is true for the average, it must be true for at least one of the sequences.
