In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n? Let $p$, $q$, $r$, and $s$ be the marks scored in papers P, Q, R, and S respectively. Since the maximum marks for each paper is 100, we have $0 \le p, q, r, s \le 100$. The total marks is $p+q+r+s$. The maximum possible total marks is $4 \times 100 = 400$. 99% of the maximum marks is $0.99 \times 400 = 396$. We need to find the number of integer solutions to the equation $p+q+r+s = 396$, subject to the constraint $0 \le p, q, r, s \le 100$.

Without the upper bound restriction, the number of non-negative integer solutions is given by $\binom{396+4-1}{4-1} = \binom{399}{3}$. However, we have the constraint $p, q, r, s \le 100$. Let $p' = 100 - p$, $q' = 100 - q$, $r' = 100 - r$, $s' = 100 - s$. Then $0 \le p', q', r', s' \le 100$. $p+q+r+s = 396$ $(100-p')+(100-q')+(100-r')+(100-s') = 396$ $400 - (p'+q'+r'+s') = 396$ $p'+q'+r'+s' = 400-396 = 4$ The number of non-negative integer solutions to $p'+q'+r'+s'=4$ is $\binom{4+4-1}{4-1} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. Thus, $n = 35$.

Final Answer: The final answer is $\boxed{35}$