Answer: $1326$

Pick any contestant $x$. Suppose he has played against $y_1, y_2, \dots, y_{100}$ (call these Group Y contestants). Denote the contestants who have not played against $x$ by $z_1, z_2, \dots, z_m$ (call these Group Z contestants).

A Group Y contestant (who has played against $x$) has exactly $50$ common opponents with $x$ (who must be Group Y contestants), and so has played against exactly $100 - 1 - 50 = 49$ Group Z contestants. On the other hand, since a Group Z contestant has not played against $x$, he has exactly $4$ common opponents with $x$ (who must be Group Y contestants). Thus the number of games between a Group Y contestant and a Group Z contestant is equal to $100 \times 49$ as well as $4m$. It follows that these numbers are equal, and so
$$n = 1 + 100 + m = 1 + 100 + \frac{100 \times 49}{4} = 1326.$$
The scenario described in the question is indeed possible. Note that the answer is actually $\binom{52}{2}$. To construct the scenario, label the contestants by $2$-element subsets of $\{1, 2, \dots, 52\}$, and let two contestants play a game if and only if their corresponding subsets have a common element. It is easy to check that all given conditions are satisfied.