Let $m = 7$ and $n = 10$. We now claim that the minimum possible value is $m-1$ and this is achieved by letting $a_i = m^i$. Now, to show that this is indeed the minimum, we assume without loss of generality that $a_n$ is the largest amongst all $a_i$. Now,
$$\sum_{i=1}^{n} \left\lfloor \frac{ma_i}{a_i + a_{i+1}} \right\rfloor \ge \left\lfloor \frac{ma_1}{a_1 + a_2} \right\rfloor + \left\lfloor \frac{ma_n}{a_n + a_1} \right\rfloor \ge \left\lfloor \frac{ma_1}{a_1 + a_n} \right\rfloor + \left\lfloor \frac{ma_n}{a_n + a_1} \right\rfloor \ge \left\lfloor \frac{m(a_1 + a_n)}{a_n + a_1} \right\rfloor - 1 \ge m - 1$$
We just use the fact that $\lfloor x \rfloor + \lfloor y \rfloor \ge \lfloor x + y \rfloor - 1$.

This tells us that the problem is indeed true in much higher generality:
Let $m, n \ge 2$ be positive integers. Given that $a_1, \dots, a_n$ are positive real numbers, determine the smallest possible value of
$$\sum_{i=1}^{n} \left\lfloor \frac{ma_i}{a_i + a_{i+1}} \right\rfloor$$
where we define $a_{n+1} = a_1$.

We first observe that if $a_i \ge a_{i+1}$, then
$$\left\lfloor \frac{7a_i}{a_i + a_{i+1}} \right\rfloor \ge \left\lfloor \frac{7a_i}{2a_i} \right\rfloor \ge 3.$$
Thus the sequence (taken cyclically) decreases at most once, otherwise we are done. WLOG $a_1 < a_2 < \dots < a_{10}$.
Make the substitution $a_{i+1}/a_i = x_i$. We have $x_i > 1$ for $1 \le i \le 9$, and $x_{10} = 1/x_1 x_2 \dots x_9$. The expression becomes
$$\sum_{i=1}^{9} \left\lfloor \frac{7}{1+x_i} \right\rfloor + \left\lfloor \frac{7x_1 x_2 \dots x_9}{x_1 x_2 \dots x_9 + 1} \right\rfloor.$$
If we have $x_i \ge 6$ for any $1 \le i \le 9$, then we get
$$\begin{aligned} \left\lfloor \frac{7x_1x_2\dots x_9}{x_1x_2\dots x_9+1} \right\rfloor &\ge 6, \\ \frac{7x_1x_2\dots x_9}{x_1x_2\dots x_9+1} &\ge 6x_1x_2\dots x_9+6 \\ x_1x_2\dots x_9 &\ge 6 \end{aligned}$$

and we are done.
If not, $x_i < 6$ for all $1 \le i \le 9$. But then, for $1 \le i \le 9$,
$$\left\lfloor \frac{7}{1 + x_i} \right\rfloor \ge 1,$$
which means the expression is at least 9, which is greater than our bound! ☐

(uses $m = 7, n = 10$ critically) If at most 4 of the $z_i$'s are zero, then at least 6 of them are $\ge 1$, and thus the sum is at least 6.
If not, then at least five of them are 0. Suppose $z_1 = \cdots = z_5 = 0$ without loss of generality. Note that $z_i = 0$ implies
$$\frac{7}{1 + x_i} < 1 \implies x_i > 6,$$
so that $x_1, \dots, x_5 > 6$. Thus $x_1x_2x_3x_4x_5 > 6^5$, which means $x_6x_7x_8x_9x_{10} < 1/6^5$. Assuming WLOG $x_6 = \min\{x_6, x_7, x_8, x_9, x_{10}\}$, this implies $x_6 < 1/6$. However, then we have
$$z_6 \ge \left\lfloor \frac{7}{1 + \frac{1}{6}} \right\rfloor = 6,$$
and the conclusion follows. ☐