THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

The required minimum is $1$. To show that $1$ is an upper bound, let $n$ be a positive integer, and let $a_1,\dots ,a_n$ be positive real numbers such that $a_1+\cdots +a_n\le \pi$. If $a_1\ge \pi /2$, then the sum in question is non-positive, so let $a_1<\pi /2$, and let $m$ be the largest positive integer such that $a_1+\cdots +a_m<\pi /2$. Then $$\sum _{k=1}^n{a}_k\cos (a_1+\cdots +a_k)\le \sum _{k=1}^m{a}_k\cos (a_1+\cdots +a_k)\le {\int }_0^{\pi /2}\cos xdx=1,$$ since the sum in the middle is the lower Darboux-Riemann sum of the cosine, corresponding to the subdivision $0<a_1<a_1+a_2<\cdots <a_1+\cdots +a_m<\pi /2$, and the cosine is decreasing on $[0,\pi /2]$.

To show that $1$ is the least upper bound, for every positive integer $n$, let $a_1=\cdots =a_n=\pi /(2n)$. Since $a_1+\cdots +a_n=\pi /2$ and $$\underset{n\to \infty }{\lim }\sum _{k=1}^n{a}_k\cos (a_1+\cdots +a_k)=\underset{n\to \infty }{\lim }\frac{\pi }{2n}\sum _{k=1}^n\cos \frac{k\pi }{2n}={\int }_0^{\pi /2}\cos xdx=1,$$ the conclusion follows.