The sum of all natural numbers $$n$$ such that $$100 < n < 200$$ and H.C.F. $$(91, n) > 1$$ is:

We are asked to consider all natural numbers $$n$$ that satisfy the double inequality $$100<n<200$$ and whose highest common factor with $$91$$ is greater than $$1$$.

First, we factorise $$91$$ so that we know which prime numbers can possibly divide both $$91$$ and $$n$$.

We have $$91 = 7 \times 13$$, and $$7$$ and $$13$$ are primes.

If the H.C.F. (also called G.C.D.) of $$91$$ and $$n$$ is to be greater than $$1$$, then $$n$$ must share at least one of the prime factors $$7$$ or $$13$$ with $$91$$. Equivalently, $$n$$ must be divisible by $$7$$ or by $$13$$ (or by both).

So we now list every multiple of $$7$$ and every multiple of $$13$$ that lies strictly between $$100$$ and $$200$$.

Multiples of $$7$$ in the required interval
We solve $$7k$$ with $$100<7k<200$$.

Dividing by $$7$$ gives $$\dfrac{100}{7}<k<\dfrac{200}{7}$$, i.e. $$14.28\dots<k<28.57\dots$$. Hence the integer values of $$k$$ are $$k=15,16,\dots ,28$$. Writing the corresponding numbers:

$$\begin{aligned} 7\times15 &= 105 \\ 7\times16 &= 112 \\ 7\times17 &= 119 \\ 7\times18 &= 126 \\ 7\times19 &= 133 \\ 7\times20 &= 140 \\ 7\times21 &= 147 \\ 7\times22 &= 154 \\ 7\times23 &= 161 \\ 7\times24 &= 168 \\ 7\times25 &= 175 \\ 7\times26 &= 182 \\ 7\times27 &= 189 \\ 7\times28 &= 196 \end{aligned}$$

So the complete list of admissible multiples of $$7$$ is
$$105,112,119,126,133,140,147,154,161,168,175,182,189,196.$$

Multiples of $$13$$ in the required interval
We solve $$13m$$ with $$100<13m<200$$.

Dividing by $$13$$ gives $$\dfrac{100}{13}<m<\dfrac{200}{13}$$, i.e. $$7.69\dots<m<15.38\dots$$. Hence the integer values of $$m$$ are $$m=8,9,\dots ,15$$. Writing the corresponding numbers:

$$\begin{aligned} 13\times8 &= 104 \\ 13\times9 &= 117 \\ 13\times10 &= 130 \\ 13\times11 &= 143 \\ 13\times12 &= 156 \\ 13\times13 &= 169 \\ 13\times14 &= 182 \\ 13\times15 &= 195 \end{aligned}$$

So the complete list of admissible multiples of $$13$$ is
$$104,117,130,143,156,169,182,195.$$

Combining the two lists
Every number in either list satisfies the H.C.F. condition, but we must beware of double counting any number that is simultaneously a multiple of $$7$$ and $$13$$.

Notice that $$182$$ appears in both lists because $$182 = 7\times26 = 13\times14$$ (since $$182$$ is itself $$91\times2$$). No other number is common to the two lists.

Sum of all admissible numbers

First sum all the multiples of $$7$$ we found:

$$\begin{aligned} &105+112+119+126+133+140+147+154+161+168+175+182+189+196 \\ &= (105+112) + (119+126) + (133+140) + (147+154) + (161+168) + (175+182) + (189+196) \\ &= 217 + 245 + 273 + 301 + 329 + 357 + 385 \\ &= 2107 \end{aligned}$$

Next sum all the multiples of $$13$$ we found:

$$\begin{aligned} &104+117+130+143+156+169+182+195 \\ &= (104+117) + (130+143) + (156+169) + (182+195) \\ &= 221 + 273 + 325 + 377 \\ &= 1196 \end{aligned}$$

Because $$182$$ lies in both lists, it has been counted twice, once in each subtotal. To obtain the correct total we subtract this overlap once.

Therefore the required sum is

$$2107 + 1196 - 182 = 3121.$$

Hence, the correct answer is Option C.