Answer: C

Solution:

Going from $n$ to $n + 1,$ each remainder $n \bmod m$ increases by $1$ unless $m \mid n + 1,$ in which case it drops from $m - 1$ to $0.$ So $$\begin{aligned} &R(n+1) - R(n) \\ &= 9 - \sum_{\substack{2 \le m \le 10 \\ m \mid n+1}} m. \end{aligned}$$

Distinct integers from $2$ through $10$ can sum to $9$ only as $\{9\},\{2,7\},\{3,6\},\{4,5\},$ or $\{2,3,4\}.$ Divisibility closure eliminates every case except $\{2,7\}:$ for example, a multiple of $9$ also has divisor $3,$ and a multiple of $6$ also has divisors $2$ and $3.$ Thus $n+1$ must be a multiple of $14$ with no other divisor in $\{2,\ldots,10\}.$

Testing the multiples $14,28,42,56,70,84,98$ in the required range leaves only $14$ and $98.$ Hence $n=13$ or $n=97,$ giving $2$ values.

Thus, the correct answer is C.