It follows from the conditions of the problem that after $x$ matches the team had $y$ wins, such that $\frac{y}{x} = \frac{n}{100}$. Then, the following has to hold: $\frac{y+1}{x+1} = \frac{n+1}{100}$.

The first equation implies $100y = nx$, and so $100$ has to divide $nx$.
The second equation implies $100y + 100 = nx + x + n + 1 \Rightarrow 100 = x + n + 1$.
Hence $x = 100 - n - 1 \le 98$. Moreover, $n = 100 - x - 1 \Rightarrow xn = 100x - x^2 - x$. Therefore, $100$ has to divide $x^2 + x$. Since $x^2 + x = x(x+1)$ and the numbers $x$ and $x+1$ are relatively prime and less than $100$, one of the numbers has to be divisible by $4$, and the other by $25$. Clearly, the smallest such number is $x = 24$. Thus, $n = 75$ and $y = 18$. It is easy to check that
$$\frac{y}{x} = \frac{18}{24} = 75\% \text{ and } \frac{y+1}{x+1} = \frac{19}{25} = 76\%$$