Why not just fix the computer?

The maximum number of roads is achieved when $100$ cities are connected in such a way that any two cities are connected by a road, then the $101$ st city is connected to any of the $100$ cities, resulting in ${100 \choose 2} + 1 = 4951$ roads.

We then prove that $4951$ roads will always connect all of the cities. Assume that there are two groups of connected cities which are not connected to each other. Let the first group have $x$ cities. The second group will then have $101-x$ cities. The first group has at most ${x \choose 2} = \frac{(x)(x-1)}{2}$ roads. Similarly, the second group has at most ${101-x \choose 2} = \frac{(101-x)(100-x)}{2}$ roads. All in all, the two groups will have $\frac{(x)(x-1)}{2} + \frac{(101-x)(100-x)}{2} = x^2 - 101x + 5050$ roads.

Now, this is a quadratic which is maximised when $|50.5 - x|$ is maximised. Since we cannot have $x = 101$ or $x=0$ (Since this will connect all of the cities), we can either have $x = 1$ or $x = 100$ . This sets the total number of roads for which the country is not connected to be $4950$ at maximum. Therefore adding one road will connect the country, so our answer is $\boxed{4951}$ .

In the worst case scenario, 100 cities will be connected in all possible ways and the last city (the 101st) will be connected directly by one of them (and indirectly, by all other cities).

Therefore, the maximum number of roads is:

$\frac {100 × 99}{2} + 1 = \boxed {4951}$