a) Suppose the contrary. Let $A_k: =a_1+2a_2+\ldots+ka_k$. \[ A_k> \frac{2k+1}4 \] Suppose that $a_1\geq 0$ or else negate all $a_i$s. We say that all $A_k>0$. If it's not true consider the first $A_k$ that is negative. Then $A_{k-1}>0$. So we have $A_k<-\frac{2k+1}4$ and $A_{k-1}>\frac{2k-1}4$. So we have $-A_{k-1}<-\frac{2k-1}4$ and by adding this to $A_k<-\frac{2k+11}4$ we get \[ ka_k<-k \] which is not possible because $a_k\geq -1$. So all $A_k>0$. Now consider $c_i$s defined below: \[ c_n=\frac 1n=\frac 1n-0 \] \[ c_{n-k}=\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1} (k\geq 1) \] Then by the right hand side of these equalities we can see that $c_n+c_{n-1}+\ldots+c_k=\frac 1k$ for $0\leq k\leq n$. Now consider the sum $\sum_{i=1}^n c_iA_i$ the coefficent of $a_{n-k}$ in this sum is \[ \sum_{i=n-k}^n (n-k)c_i=(n-k)\sum_{i=n-k}^n c_i=(n-k)\frac 1{n-k}=1 \] So the sum is equal to $0$ but every term of the sum is positive and that cannot happen! b) Just take $a_1=\frac 34$ and $a_2=\frac 14$ and $a_i=(-1)^i$ for every $i\geq 3$. Then we can check that $A_1=\frac 34$ and $A_2=\frac 54$ and for $i\geq 3$ using induction we can see that \[ A_i=(-1)^i\frac{2k+1}4 \]

This reminds me of IMO 1997 Problem 3

Yes, Arne. I thought of that the moment I saw this problem. EDIT: This problem was proposed by Radu Gologan and Dan Schwarz and was inspired by the problem that I had given in National Olympiad, 12-grade.

Nima Ahmadi Pour wrote: We say that all $A_k>0$. If it's not true consider the first $A_k$ that is negative. Then $\boxed{A_{k-1}>0}$. I don't understand here.

Nima Ahmadi Pour wrote: b) Just take $a_{1}=\frac{3}{4}$ and $a_{2}=\frac{1}{4}$ and $a_{i}=(-1)^{i}$ for every $i\geq 3$. So if $n$ is even then $a_{1}+...+a_{n}$ is not $0$...

Can anybody fix what I said?

Nobody?

For odd $n$ take the numbers from the solution of Nima Ahmadi Pour For even $n$ take :$a_1=1; a_2=\frac{1}{8}; a_3=-1; a_4=-\frac{1}{8}; a_i=(-1)^{i+1}$ for $i\in \lbrace 5,6,...,n\rbrace$.