We use an induction on $n$.

a. For $n = 6$, put $N_6 = 2 \cdot 3 \cdot 7 \cdot 43 \cdot 1807 = 1806 \cdot 1807$. Since $1807 = 13 \cdot 139$, $N_6$ is divisible by 6 distinct primes.
Moreover, since
$$\frac{1}{m} = \frac{1}{m+1} + \frac{1}{m(m+1)},$$
we have

$$\begin{align*} 1 &= \frac{1}{2} + \frac{1}{2} \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1806}, \quad (42 \cdot 43 = 1806) \\ &= \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1807} + \frac{1}{N_6}. \end{align*}$$
Hence, by multiplying both sides of above equality by $N_6$, we get
$$N_6 = 1 + \frac{N_6}{2} + \frac{N_6}{3} + \frac{N_6}{7} + \frac{N_6}{43} + \frac{N_6}{1807},$$
where each term of the right hand side is a divisor of $N_6$.

b. Suppose there is an $n$-good number $N_n$. Put
$$N_{n+1} = N_n(N_n + 1).$$
Then
$$\begin{align*} N_{n+1} &= (1 + x_2 + \cdots + x_n)(N_n + 1) \\ &= 1 + N_n + x_2(N_n + 1) + \cdots + x_n(N_n + 1). \end{align*}$$
Hence $N_{n+1}$ is a sum of $n+1$ distinct divisors.
Since $(N_n, N_n + 1) = 1$, prime divisors of $N_n + 1$ are different from those of $N_n$. Since $N_n + 1$ has at least one prime divisor, and since $N_n$ has at least $n$ distinct prime divisors, $N_{n+1}$ has at least $n+1$ distinct prime divisors. Therefore $N_{n+1}$ is a $(n+1)$-good number. $\square$