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 $a_1+\dots+a_n \le \pi$. 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 $a_1 + \dots + a_m < \pi/2$. Then
$$\sum_{k=1}^{n} a_k \cos(a_1 + \dots + a_k) \le \sum_{k=1}^{m} a_k \cos(a_1 + \dots + a_k) \le \int_{0}^{\pi/2} \cos x \, dx = 1,$$
since the sum in the middle is the lower Darboux-Riemann sum of the cosine, corresponding to the subdivision $0 < a_1 < a_1 + a_2 < \dots < a_1 + \dots + a_m < \pi/2$, and the cosine is decreasing on $[0, \pi/2]$.

To show that $1$ is the least upper bound, for every positive integer $n$, let $a_1 = \dots = a_n = \pi/(2n)$. Since $a_1 + \dots + a_n = \pi/2$ and
$$\lim_{n \to \infty} \sum_{k=1}^{n} a_k \cos(a_1 + \dots + a_k) = \lim_{n \to \infty} \frac{\pi}{2n} \sum_{k=1}^{n} \cos \frac{k\pi}{2n} = \int_{0}^{\pi/2} \cos x \, dx = 1,$$
the conclusion follows.