Clearly the tournament can be run in $99$ days if we assign, on day $k$ (where $1 \le k \le 99$),

* player $k$ to be 'white';
* players $k + 1$, $k + 2$, $\ldots$, $100$ to be 'black'; and
* all other players to be 'idle'.

In this way player $i$ and player $j$ (where $1 \le i < j \le 100$) would have met exactly once on day $i$.

Now suppose the number of days is less than $99$, and we would derive a contradiction. Note that we can assign a real number $x_i$ to each player $i$, such that
(1) at least one $x_i$ is nonzero;
(2) the sum of all $x_i$ is zero; and
(3) the sum of the $x_i$'s of all 'white' players on each day is zero.

This is because (2) and (3) represent a homogeneous system of fewer than $100$ linear equations in $100$ unknowns, and hence must have a nonzero solution (i.e. condition (1)). It follows that
$$0 = (x_1 + x_2 + \cdots + x_{100})^2 = (x_1^2 + x_2^2 + \cdots + x_{100}^2) + 2 \sum_{1 \le i < j \le 100} x_i x_j. \quad (4)$$
Since any two players have met exactly once, we have
$$\sum_{1 \le i < j \le 100} x_i x_j = \sum_k \left( \sum_{w \in W_k} x_w \sum_{b \in B_k} x_b \right)$$
where $W_k$ and $B_k$ denote the set of 'white' and 'black' players on day $k$ respectively. Since the sum of the $x_i$'s of all 'white' players on each day is zero, the above summation is equal to $0$. On the other hand, $x_1^2 + x_2^2 + \cdots + x_{100}^2 > 0$ as at least one $x_i$ is nonzero. This gives a contradiction to (4).