Step 1: Express the general term of the sum in a closed form.

Let the general term be $T_r = 7 \overbrace{5 \cdots 5}^r 7$. This is an $(r+2)$-digit number. We can express this number algebraically: $T_r = 7 \cdot 10^{r+1} + 5 \cdot 10^r + 5 \cdot 10^{r-1} + \cdots + 5 \cdot 10^1 + 7 \cdot 10^0$ $T_r = 7 \cdot 10^{r+1} + 7 + 5(10^r + 10^{r-1} + \cdots + 10)$ The series in the parenthesis is a geometric progression with first term $a=10$, common ratio $R=10$, and $r$ terms. The sum is $10 \frac{10^r-1}{10-1} = \frac{10}{9}(10^r-1)$. So, $T_r = 7 \cdot 10^{r+1} + 7 + 5 \cdot \frac{10}{9}(10^r-1) = 7 \cdot 10^{r+1} + 7 + \frac{50}{9}(10^r-1)$. To simplify, we take a common denominator of 9: $T_r = \frac{9(7 \cdot 10^{r+1} + 7) + 50(10^r-1)}{9}$ $T_r = \frac{63 \cdot 10^{r+1} + 63 + 50 \cdot 10^r - 50}{9}$ $T_r = \frac{630 \cdot 10^r + 50 \cdot 10^r + 13}{9}$ $T_r = \frac{680 \cdot 10^r + 13}{9}$

Step 2: Calculate the sum S.

The sum is given by $S=77+757+7557+\cdots+7 \overbrace{5 \cdots 5}^{98}7$. The terms correspond to $r=0, 1, 2, \dots, 98$. There are $98-0+1 = 99$ terms in the sum. $S = \sum_{r=0}^{98} T_r = \sum_{r=0}^{98} \frac{680 \cdot 10^r + 13}{9}$ $S = \frac{1}{9} \left( \sum_{r=0}^{98} 680 \cdot 10^r + \sum_{r=0}^{98} 13 \right)$ The first part is a sum of a geometric series: $\sum_{r=0}^{98} 680 \cdot 10^r = 680 \sum_{r=0}^{98} 10^r = 680 \left( \frac{10^{99}-1}{10-1} \right) = \frac{680}{9}(10^{99}-1)$. The second part is the sum of a constant: $\sum_{r=0}^{98} 13 = 13 \times 99$. Substituting these back into the expression for S: $S = \frac{1}{9} \left( \frac{680}{9}(10^{99}-1) + 13 \times 99 \right)$ $S = \frac{1}{81} \left( 680(10^{99}-1) + 13 \times 99 \times 9 \right)$ $S = \frac{1}{81} \left( 680 \cdot 10^{99} - 680 + 13 \times 891 \right)$ $S = \frac{1}{81} \left( 680 \cdot 10^{99} - 680 + 11583 \right)$ $S = \frac{680 \cdot 10^{99} + 10903}{81}$

Step 3: Use the given form of S to find m and n.

The problem states that $S=\frac{7 \overbrace{5 \cdots 5}^{99}7+m}{n}$. The number $7 \overbrace{5 \cdots 5}^{99}7$ is the term $T_r$ with $r=99$. So, $7 \overbrace{5 \cdots 5}^{99}7 = T_{99} = \frac{680 \cdot 10^{99} + 13}{9}$. Substituting this into the given form of S: $S = \frac{\frac{680 \cdot 10^{99} + 13}{9} + m}{n} = \frac{680 \cdot 10^{99} + 13 + 9m}{9n}$

Step 4: Equate the two expressions for S and solve for m and n.

We have two expressions for S: $S = \frac{680 \cdot 10^{99} + 10903}{81} \quad \text{and} \quad S = \frac{680 \cdot 10^{99} + 13 + 9m}{9n}$ Equating them: $\frac{680 \cdot 10^{99} + 10903}{81} = \frac{680 \cdot 10^{99} + 13 + 9m}{9n}$ Let $A = 680 \cdot 10^{99}$. $\frac{A + 10903}{81} = \frac{A + 13 + 9m}{9n}$ $9n(A + 10903) = 81(A + 13 + 9m)$ Dividing by 9: $n(A + 10903) = 9(A + 13 + 9m)$ $nA + 10903n = 9A + 117 + 81m$ $(n-9)A = 117 + 81m - 10903n$ Since $A = 680 \cdot 10^{99}$ is an extremely large integer and $m, n$ are natural numbers less than 3000, the right-hand side is a relatively small integer. For this equality to hold, the coefficient of A must be zero. Therefore, $n-9=0 \implies n=9$. This also implies the right-hand side must be zero: $117 + 81m - 10903n = 0$ Substitute $n=9$: $117 + 81m - 10903(9) = 0$ $117 + 81m - 98127 = 0$ $81m = 98127 - 117$ $81m = 98010$ $m = \frac{98010}{81} = 1210$. We have $m=1210$ and $n=9$. Both are natural numbers less than 3000, as required.

Step 5: Calculate the final value.

The question asks for the value of $m+n$. $m+n = 1210 + 9 = 1219$.

The final answer is 1219.