Two simple lemmas are needed to prove the problem.

Lemma. For all integers $k > 1$, if $x_1, x_2, \dots, x_k$ are integers with
$$\gcd(x_1, \dots, x_k) = 1,$$
then there are integers $a_1, \dots, a_k$ such that
$$a_1x_1 + \dots + a_kx_k = 1.$$

Proof. For $k = 2$ the statement of the lemma is Bézout's Lemma for relatively prime numbers, that is well-known. Now we use induction on $k$. Knowing that the lemma is true for $k = n$, for numbers $x_1, \dots, x_{n+1}$ that are relatively prime, we have
$$\text{gcd}(x_1, x_2, \dots, x_{n-1}, \text{gcd}(x_n, x_{n+1})) = 1.$$
Now applying the lemma for $k = n$, there are numbers $a_1, \dots, a_n$ such that
$$a_1x_1 + a_2x_2 + \dots + a_{n-1}x_{n-1} + a_n\text{gcd}(x_n, x_{n+1}) = 1$$
Using the Bézout's Lemma, there are $b_n, b_{n+1}$ such that
$$\text{gcd}(x_n, x_{n+1}) = b_n x_n + b_{n+1} x_{n+1}.$$
Set $y_i = a_i$ for $i < n$, $y_n = a_n b_n$ and $y_{n+1} = a_n b_{n+1}$ to get
$$y_1 x_1 + \dots + y_{n+1} x_{n+1} = 1.$$
Which is the statement of the lemma for $k = n + 1$. So the lemma is proved using induction.

Lemma. Set $L = \text{lcm}(1, 2, \dots, n)$, then $\text{gcd}(\frac{L}{m+1}, \frac{L}{m+2}, \dots, \frac{L}{n}) = 1$ where $m = \lfloor \frac{n}{2} \rfloor$.

Proof. First note that $\text{lcm}(1, 2, \dots, n) = \text{lcm}(m+1, m+2, \dots, n)$. That is because for all $1 \le k \le m$, the number $s = \lceil \log_2(\frac{m+1}{k}) \rceil$ is an integer such that
$$2^s \cdot k \in \{m+1, m+2, \dots, n\}.$$
Now assume that
$$d = \text{gcd} \left( \frac{L}{m+1}, \frac{L}{m+2}, \dots, \frac{L}{n} \right).$$
For all $m+1 \le i \le n$ we have
$$d \mid \frac{L}{i} \implies \exists k; d \cdot k = \frac{L}{i} \implies i \cdot k = \frac{L}{d} \implies i \mid \frac{L}{d}.$$
Which implies
$$L = \text{lcm}(m+1, m+2, \dots, n) \mid \frac{L}{d} \implies d = 1.$$

Back to the problem, according to both lemmas, since
$$\text{gcd} \left( \frac{L}{m+1}, \frac{L}{m+2}, \dots, \frac{L}{n} \right) = 1,$$
we obtain there are some integers $x_m, x_{m+1}, \dots, x_{n-1}$ such that
$$x_m \cdot \frac{L}{m+1} + x_{m+1} \cdot \frac{L}{m+2} + \dots + x_{n-1} \cdot \frac{L}{n} = 1.$$
Now consider a natural number $t$ such that $x_m + t \cdot \frac{L}{m+2} > 0$ and set
$$\begin{cases} a_m = x_m + t \cdot \frac{L}{m+2}, \\ a_{m+1} = x_{m+1} - t \cdot \frac{L}{m+1}, \\ a_i = x_i, \end{cases} \quad \forall i > m+1$$
to get
$$\frac{a_m}{m+1} + \frac{a_{m+1}}{m+2} + \dots + \frac{a_{n-1}}{n} = \frac{1}{\text{lcm}(1, 2, \dots, n)}.$$
Since $a_m = x_m + t \cdot \frac{L}{m+2} > 0$, this is the desired equation for $m = \lfloor \frac{n}{2} \rfloor$.