To solve the problem, we need to find the sum of all natural numbers \( n \) such that \( 100 < n < 200 \) and \( \text{H.C.F.}(91, n) > 1 \).
### Step 1: Factor the number 91
First, we factor 91 to find its prime factors:
\[
91 = 7 \times 13
\]
This means that any number \( n \) for which \( \text{H.C.F.}(91, n) > 1 \) must be divisible by either 7 or 13.
### Step 2: Find numbers divisible by 7
Next, we find all natural numbers \( n \) in the range \( 100 < n < 200 \) that are divisible by 7.
The smallest number greater than 100 that is divisible by 7 is:
\[
105 \quad (7 \times 15)
\]
The largest number less than 200 that is divisible by 7 is:
\[
196 \quad (7 \times 28)
\]
The numbers divisible by 7 in this range are:
\[
105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196
\]
### Step 3: Calculate the sum of numbers divisible by 7
We can use the formula for the sum of an arithmetic series:
\[
S = \frac{n}{2} \times (a + l)
\]
where:
- \( n \) is the number of terms,
- \( a \) is the first term,
- \( l \) is the last term.
Here, the first term \( a = 105 \), the last term \( l = 196 \), and the number of terms \( n \) can be calculated as:
\[
n = \frac{196 - 105}{7} + 1 = \frac{91}{7} + 1 = 13 + 1 = 14
\]
Thus, the sum \( S_A \) of numbers divisible by 7 is:
\[
S_A = \frac{14}{2} \times (105 + 196) = 7 \times 301 = 2107
\]
### Step 4: Find numbers divisible by 13
Now, we find all natural numbers \( n \) in the range \( 100 < n < 200 \) that are divisible by 13.
The smallest number greater than 100 that is divisible by 13 is:
\[
104 \quad (13 \times 8)
\]
The largest number less than 200 that is divisible by 13 is:
\[
195 \quad (13 \times 15)
\]
The numbers divisible by 13 in this range are:
\[
104, 117, 130, 143, 156, 169, 182, 195
\]
### Step 5: Calculate the sum of numbers divisible by 13
Using the same formula for the sum:
Here, the first term \( a = 104 \), the last term \( l = 195 \), and the number of terms \( n \) can be calculated as:
\[
n = \frac{195 - 104}{13} + 1 = \frac{91}{13} + 1 = 7 + 1 = 8
\]
Thus, the sum \( S_B \) of numbers divisible by 13 is:
\[
S_B = \frac{8}{2} \times (104 + 195) = 4 \times 299 = 1196
\]
### Step 6: Find numbers divisible by both 7 and 13 (i.e., 91)
Now, we find all natural numbers \( n \) in the range \( 100 < n < 200 \) that are divisible by 91.
The smallest number greater than 100 that is divisible by 91 is:
\[
91 \times 2 = 182
\]
The largest number less than 200 that is divisible by 91 is:
\[
91 \times 2 = 182
\]
The only number divisible by both 7 and 13 in this range is:
\[
182
\]
### Step 7: Calculate the sum of numbers divisible by both 7 and 13
The sum \( S_N \) of numbers divisible by both 7 and 13 is:
\[
S_N = 182
\]
### Step 8: Use the principle of inclusion-exclusion
Now, we can find the total sum \( S \) of numbers \( n \) such that \( \text{H.C.F.}(91, n) > 1 \):
\[
S = S_A + S_B - S_N = 2107 + 1196 - 182 = 3121
\]
### Final Answer
Thus, the sum of all natural numbers \( n \) such that \( 100 < n < 200 \) and \( \text{H.C.F.}(91, n) > 1 \) is:
\[
\boxed{3121}
\]
