To find natural numbers $N$ that can be expressed as the sum of three of its distinct factors $p$, $q$, and $r$, we need to analyze the mathematical relationship between a number and its divisors.

Mathematical Framework

Let the factors be $p, q, r$. Since they are factors of $N$, they can be written as:

$$p = \frac{N}{a}, \quad q = \frac{N}{b}, \quad r = \frac{N}{c}$$

where $a, b, c$ are distinct integers greater than or equal to 1.

The given condition is:

$$N = \frac{N}{a} + \frac{N}{b} + \frac{N}{c}$$

Dividing both sides by $N$ (since $N$ is a natural number, $N \neq 0$):

$$1 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$

Finding the Possible Values for $a, b, c$

We need to find sets of three distinct positive integers whose reciprocals sum to 1. Without loss of generality, assume $a < b < c$.

If $a = 1$, the sum would already be 1, making $1/b$ and $1/c$ zero, which is impossible for finite $b$ and $c$.
If $a = 2$, we need:
$$\frac{1}{b} + \frac{1}{c} = 1 - \frac{1}{2} = \frac{1}{2}$$
- If $b=3$, then $1/c = 1/2 - 1/3 = 1/6$. So, $c=6$.
- If $b=4$, then $1/c = 1/2 - 1/4 = 1/4$. But $b$ and $c$ must be distinct, so this is invalid.
- If $b > 4$, the sum $1/b + 1/c$ will be less than $1/2$.
If $a = 3$, the maximum sum would be $1/3 + 1/4 + 1/5 = 47/60$, which is less than 1.

Thus, the only possible set for $\{a, b, c\}$ is $\{2, 3, 6\}$.

Translating Back to $N$

Since $a, b, c$ are $\{2, 3, 6\}$, the factors $p, q, r$ must be:

$$p = \frac{N}{2}, \quad q = \frac{N}{3}, \quad r = \frac{N}{6}$$

For $N/2, N/3,$ and $N/6$ to be integers (factors), $N$ must be a multiple of 6.

Identifying Numbers Between 50 and 100

We are looking for multiples of 6 in the range $50 < N < 100$.

Multiples of 6	Is it between 50 and 100?
$6 \times 9 = 54$	Yes
$6 \times 10 = 60$	Yes
$6 \times 11 = 66$	Yes
$6 \times 12 = 72$	Yes
$6 \times 13 = 78$	Yes
$6 \times 14 = 84$	Yes
$6 \times 15 = 90$	Yes
$6 \times 16 = 96$	Yes

All multiples of 6 from $6 \times 9$ to $6 \times 16$ fall within the specified range.

Verification of an Example

Let's take $N = 54$:

Factors of 54 are 1, 2, 3, 6, 9, 18, 27, 54.
$N/2 = 27$, $N/3 = 18$, $N/6 = 9$.
Sum: $27 + 18 + 9 = 54$.
This matches $N$.

Conclusion

The number of such values is the count of multiples of 6 between 50 and 100:

$$16 - 9 + 1 = 8$$

There are 8 such numbers.