To solve the problem, we need to find the sum of all positive integers $n$ not exceeding $100,000,000$ such that for every divisor $d$ of $n$, the expression $d + \frac{n}{d}$ is prime.

Analyzing the Condition $d + \frac{n}{d}$ is Prime

For any $n$, the divisors $d$ and $\frac{n}{d}$ pair up, and $d + \frac{n}{d}$ should be a prime number. This implies the structure of $n$ must be such that the sum $d + \frac{n}{d}$ is prime for all divisors $d$.

Let us first investigate a number with exactly two distinct prime factors, i.e., $n = p_1^{a_1} \times p_2^{a_2}$.

Given the condition, $d + \frac{n}{d}$ must be prime for all divisors $d$. Consider the following cases:

1. Case $n = p^k$ for some prime $p$:

   • Here, the divisors of $n$ are $1, p, p^2, \ldots, p^k$.
   • The sums are $1 + p^k$, $p + p^{k-1}$, $p^2 + p^{k-2}$, ..., all of which must be prime.
   • This situation generally requires $n$ to be of a special form.

2. Case $n = p_1 \times p_2$ where $p_1$ and $p_2$ are distinct primes:

   • Here, divisors are $1, p_1, p_2, p_1p_2$.
   • The sums are $1 + p_1p_2$, $p_1 + p_2$, $p_2 + p_1$, and $p_1p_2 + 1$, which are unlikely to all be primes.

The analysis shows that the number must be of the form $p^k$ where $p$ is a prime number. We need to check which powers $p^k$ satisfy the given condition.

Determining Valid Values of $p^k$

For the simplest case:

• $n = p$ where $p$ is prime, we have divisors $1$ and $p$, and $1 + p$ must be prime. This condition trivially holds if $p = 2$ (since $1 + 2 = 3$ is prime).

To explore higher powers:

• If $n = p^2$, the divisors are $1, p, p^2$.
• $d + \frac{p^2}{d}$ gives $1 + p^2$, $p + p$, and $p^2 + 1$. $2p$ must be prime, which is impossible unless $p = 2$.

Thus, $n = 2^k$ is a suitable candidate.

Check the Condition for $n = 2^k$

For $n = 2^k$, divisors are $1, 2, 2^2, \ldots, 2^k$:

• $1 + 2^k$, $2 + 2^{k-1}$, ..., all need to be primes.

Specifically:

• For $k = 1$, $2$, $4$, etc., the conditions work well.

Sum All Valid $2^k$

To satisfy the condition, $n = 2^k$, where $2^k \leq 100,000,000$.

Find the maximum $k$: $2^k \leq 100,000,000 \implies k = \left\lfloor \log_2(100,000,000) \right\rfloor = 26$ So $k$ ranges from 0 to 26.

Sum: $\text{Sum} = 2^0 + 2^1 + \cdots + 2^{26} = 2^{27} - 1 = 134217727$

Therefore, the sum of all such numbers $n$ is $\boxed{134217727}$.