Solution:
Let $f(u)$ denote the sum of all nice divisors of $u$. Note that $f(u) \leq \sigma(u)$, where $\sigma(u)$ denotes the sum of all the divisors of $u$, including $1$ and $u$.

We also note the following facts about the $\sigma$ function:

1. $\sigma$ is multiplicative: if $a$ and $b$ are relatively prime (share no prime factors), then $\sigma(ab) = \sigma(a) \sigma(b)$.
2. If $p$ is prime, then $\sigma(p^{a}) = 1 + p + p^{2} + \cdots + p^{a} = \frac{p^{a+1} - 1}{p - 1} = p^{a} \left(\frac{p - 1 / p^{a}}{p - 1}\right) < p^{a} \left(\frac{p}{p - 1}\right)$.

To solve the problem, let us consider a few cases.

- Suppose $n$ is prime. Then ask the Oracle to compute $f(n)$. Clearly $f(n) \geq n$ if and only if $n$ is nice.

- Suppose $n$ is a power of a prime, so $n = p^{a}$ for some prime $p$, with $a > 1$. Again, ask the Oracle to compute $f(n)$. We claim again that $f(n) \geq n$ if and only if $n$ is nice. Clearly if $n$ is nice, then $f(n) > n$. But if $n$ is naughty, then $f(n) = f(p^{a-1}) \leq \sigma(p^{a-1}) < p^{a-1} \left(\frac{p}{p-1}\right) \leq 2 p^{a-1} \leq p^{a}$.

- Now suppose that $n$ has at least two prime factors, so we can write $n = p^{a} q^{b} R$, where $p, q$ are the two smallest primes found in the factorization of $n$, $a, b > 0$, and $R$ is an integer whose prime factorization does not contain $p$ or $q$ (of course, $R$ could equal $1$). Note that $R$ contains at most $5$ different primes, since the product of the first $8$ primes is larger than $1$ million. We will modify the previous strategy, where we examined a subset of the divisors of $n$ that were less than $n$ and whose sum was bounded below $n$; this allowed us to focus on $n$ and determine whether it is nice or naughty.

The subset of the divisors that we seek is the set $D$ of divisors of $n$ that are multiples of $p^{a} q^{b}$; in other words, $D := \{ p^{a} q^{b} r : r \text{ divides } R \}$. We claim that the sum of the elements of $D$ is less than $2n$, and since $p^{a} q^{b} R = n \in D$, we conclude that $n$ is nice if and only if the sum of the nice members of $D$ is greater than or equal to $n$ (this follows because our claim implies that the members of $D$ that are not $n$ have a sum that is strictly less than $n$).

Two things remain: to verify the claim that the sum of all the members of $D$ is less than $2n$, and we need to know how to get the Oracle to compute the sum of the nice members of $D$ with at most four questions.

1. The sum of the members of $D$ is equal to $p^{a} q^{b} \sigma(R)$. By fact #2 above,
$$\sigma(R) < R \left(\frac{p_{1}}{p_{1} - 1}\right) \left(\frac{p_{2}}{p_{2} - 1}\right) \cdots$$
where the $p_{i}$ range over the distinct prime factors of $R$. We know that there are at most five such prime factors, and this product is bounded above by
$$\left(\frac{5}{4}\right) \left(\frac{7}{6}\right) \left(\frac{11}{10}\right) \left(\frac{13}{12}\right) \left(\frac{17}{16}\right) = \frac{7 \cdot 11 \cdot 13 \cdot 17}{2^{10 \cdot 9}} = \frac{1001 \cdot 17}{1024 \cdot 9},$$
which is clearly less than $2$.

2. Finally, let the "universal set" be the divisors of $n$, and let $P$ and $Q$ denote the divisors of $n/p$ and $n/q$, respectively. It is easy to see that $P \cup Q$ is the complement of $D$ (since membership in $P$ guarantees that the exponent of $p$ is too small, and membership in $Q$ does the same for $q$). By the inclusion-exclusion principle, the members of $D$ can be obtained by starting with the universal set (all divisors of $n$), removing members of $P$, removing members of $Q$, and then adding back in the members of $P \cap Q$, which just consists of the divisors of $n/(pq)$. Consequently, the sum of the nice members of $D$ is given by
$$f(n) - f(n/p) - f(n/q) + f(n/(pq))$$
which requires just four questions to compute!