Let $p, q \in \mathbb{N}^*$, $(p, q) = 1$ so that $\frac{a}{c} = \frac{b}{d} = \frac{c}{e} = \frac{p}{q}$. Obviously, $p < q$. Since $a, b$ and $c$ are divisible by $p$ and $c, d, e$ are divisible by $q$, there exists $m \in \mathbb{N}^*$ so that $c = mpq$.
Let us find the minimal value of the difference $e - a$, for given $p, q$. This is obtained when $a, b, c$ are consecutive multiples of $p$ and $c, d, e$ are consecutive multiples of $q$, that is $a = mpq - 2p$ and $e = mpq + 2q$. From $\frac{c}{e} = \frac{mpq}{mpq+2q} = \frac{p}{q}$ follows $m(q-p) = 2$, hence $m \in \{1, 2\}$.
Condition $n \le a < e \le 2n \le 2a$ implies $2a \ge e$, that is $2mpq - 4p \ge mpq + 2q$, or $mpq \ge 4p + 2q$. (*)
If $m = 1$, then $q - p = 2$, hence $q = p + 2$. Relation (*) yields $(p-2)^2 \ge 8$, whence $p \ge 5$. For $p = 5$ and $q = 7$ we get $a = 25$, $b = 30$, $c = 35$, $d = 42$, $e = 49$ and, since $n \le a < e \le 2n$, $n = 25$.
If $m = 2$, then $q - p = 1$, so $q = p + 1$. Relation (*) yields $(p-1)^2 \ge 2$, whence $p \ge 3$. For $p = 3$ and $q = 4$ we get $a = 18$, $b = 21$, $c = 24$, $d = 28$, $e = 32$ and, since $n \le a < e \le 2n$, $n \in \{16, 17, 18\}$. Therefore, $n_{\min} = 16$.