Medals in all

Suppose that there are $x_j$ medals still to be awarded at the end of day $j$ . Then $x_j \; = \; x_{j-1} - j - \tfrac17(x_{j-1} - j) \; = \; \tfrac67(x_{j-1} - j) \hspace{2cm} 1 \le j \le n$ and hence $\big(\tfrac76\big)^jx_j \; = \; \big(\tfrac76\big)^{j-1}x_{j-1} - j\big(\tfrac76\big)^{j-1} \hspace{2cm} 1 \le j \le n$ so that $\big(\tfrac76\big)^j x_j \; = \; x_0 - \sum_{k=1}^j k \big(\tfrac76\big)^{k-1} \hspace{2cm} 1 \le j \le n$ Since $x_n = 0$ , we deduce that $m \; = \; x_0 \; = \; \sum_{k=1}^n k \big(\tfrac76\big)^{k-1}$ and so $x_j \; =\; \big(\tfrac67\big)^j\sum_{k=j+1}^n k\big(\tfrac76\big)^{k-1} \hspace{2cm} 0 \le j \le n$ In particular we must have $m \; = \; x_0 \; = \; \sum_{k=1}^n k \big(\tfrac76\big)^{k-1} \; = \; 6(n-6)\big(\tfrac76\big)^n + 36$ Since $x_j$ needs to be an integer for all $0 \le j \le n$ , we need $n=6$ , in which case $m=36$ and a total of $6$ medals are awarded each day over a period of $6$ days.

The answer is $36+6 = \boxed{42}$ .

$From~the~given~constraints,~we~have~for~a,b,c......as~integers,~the~ following~equations.\\ \begin{aligned}\\ m&=7a+1...........&Left~6a~~~~&~for~Day~2.\\ 6a-2&=7b...........&Left~6b~~~~&~for~Day~3.\\ 6b-3&=7c...........&Left~6c~~~~&~for~Day~4.\\ 6c-4&=7d...........&Left~6d~~~~&~for~Day~5.\\ 6d-5&=7e...........& \color{#3D99F6}{Left~6e}~~~~&~ \color{#3D99F6}{for~Day~6}.\\ 6e-6&=7f...........&Left~6f~~~~&~for~Day~7.\\ 6f-7&=7g...........&Left~6g~~~~&~for~Day~8.\\ ~~~\\ Let~g&=1.\\ Then~f&=\dfrac{14} 6~~Not~an~integer.\\ Let~f&=1.\\ Then~f&=\dfrac{13} 6~~Not~an~integer.\\ ~~~\\ Let~e&=1.\\ Then~d&=\dfrac{12} 6=2.\\ Then~c&=\dfrac{18} 6=3.\\ Then~b&=\dfrac{24} 6=4.\\ Then~a&=\dfrac{18} 6=5.\\ Then~m&=7*5+1=36. \end{aligned} \\ n=6. \\ \therefore~m+n=36+6=\huge \color{#D61F06}{42}.$