Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

First solution. The required maximum is $n/2$ and is achieved if and only if $a_k = 2^{k-2}a_1$, $k = 2, \dots, n$, and $a_1$ is any positive real number.
To prove this, let $A_k = a_1 + \dots + a_k$, $k = 0, \dots, n-1$, where empty sums are zero, and refer to the condition in the statement, $A_k \le a_{k+1}$, $k = 0, \dots, n-1$, to write
$$\begin{align*} \sum_{k=1}^{n-1} \frac{a_k}{a_{k+1}} &= \sum_{k=1}^{n-1} \frac{A_k - A_{k-1}}{a_{k+1}} = \sum_{k=1}^{n-2} A_k \left( \frac{1}{a_{k+1}} - \frac{1}{a_{k+2}} \right) + \frac{A_{n-1}}{a_n} \\ &\le \sum_{k=1}^{n-2} a_{k+1} \left( \frac{1}{a_{k+1}} - \frac{1}{a_{k+2}} \right) + 1 = \sum_{k=1}^{n-2} \left( 1 - \frac{a_{k+1}}{a_{k+2}} \right) + 1 \\ &= n - 1 + \frac{a_1}{a_2} - \sum_{k=1}^{n-1} \frac{a_k}{a_{k+1}} \le n - \sum_{k=1}^{n-1} \frac{a_k}{a_{k+1}}. \end{align*}$$
Consequently, $\sum_{k=1}^{n-1} a_k / a_{k+1} \le n/2$, and equality holds if and only if $A_k = a_{k+1}$, $k = 1, \dots, n-1$, which is clearly the case if and only if $a_k = 2^{k-2}a_1$, $k = 2, \dots, n$, and $a_1$ is any positive real number.

Second solution. We now show by induction on $n \ge 2$ that $s_n(a_1, \dots, a_n) = a_1/a_2 + a_2/a_3 + \dots + a_{n-1}/a_n \le n/2$ for all positive real numbers $a_1, \dots, a_n$ such that $a_k \ge a_1 + \dots + a_{k-1}$, $k = 2, \dots, n$, and equality holds if and only if $a_k = 2^{k-2}a_1$, $k = 2, \dots, n$, and $a_1$ is any positive real number.
Clearly, $s_2(a_1, a_2) \le 1$ if $0 < a_1 \le a_2$, and $s_3(a_1, a_2, a_3) \le s_3(a_1, a_2, a_1+a_2) \le 3/2$ if $0 < a_1 \le a_2$ and $a_1 + a_2 \le a_3$; in both cases, equality holds if and only if the $a_k$ are as stated.
Now let $n \ge 4$, let $a_1, \dots, a_n$ be positive real numbers such that $a_k \ge a_1 + \dots + a_{k-1}$, $k = 2, \dots, n$, and write
$$\begin{align*} s_n(a_1, \dots, a_n) &\le s_n(a_1, \dots, a_{n-1}, a_1 + \dots + a_{n-1}) \\ &= s_{n-2}(a_1, \dots, a_{n-2}) + \frac{a_{n-2}}{a_{n-1}} + \frac{a_{n-1}}{a_1 + \dots + a_{n-1}}. \end{align*}$$
If we show that
$$\frac{a_{n-2}}{a_{n-1}} + \frac{a_{n-1}}{a_1 + \cdots + a_{n-1}} \le \frac{a_{n-2}}{a_1 + \cdots + a_{n-2}} + \frac{1}{2}, \quad (*)$$
then $s_n(a_1, \dots, a_n) \le s_{n-1}(a_1, \dots, a_{n-2}, a_1 + \dots + a_{n-2}) + 1/2 \le (n-1)/2 + 1/2 = n/2$, by the induction hypothesis; the cases of equality also follow from the induction hypothesis.
Finally, to establish (*), simply notice that the numerator of the right-hand member and the left-hand member is the product of $a_{n-1} - a_1 - \dots - a_{n-2} \ge 0$ and $a_{n-1}(a_{n-2} - a_1 - \dots - a_{n-3}) + 2a_{n-2}(a_1 + \dots + a_{n-2}) \ge 2a_{n-2}(a_1 + \dots + a_{n-2}) > 0$.