Denote the players by $P_1, P_2, \dots, P_{2025}$ and let the number of matches be $N$. Also, denote the set of participants of match $i$ by $\mathcal{T}_i$ for each $i = 1, 2, \dots, N$. We will count, in two different ways, the total number of triples $(P_i, \mathcal{T}_j, \mathcal{T}_k)$ such that $P_i$ participates in match $\mathcal{T}_j$ as well in match $\mathcal{T}_k$, i.e. $P_i \in \mathcal{T}_j$ and $P_i \in \mathcal{T}_k$. Denote this total number of triples by $T$.

First we count these triples by players. Suppose, for each $1 \le i \le 2025$, player $P_i$ plays in $R_i$ matches. We note that $\sum_{i=1}^{2025} R_i = 90N$. Then the number of triples $(P_i, \mathcal{T}_j, \mathcal{T}_k)$ with $P_i \in \mathcal{T}_j$ and $P_i \in \mathcal{T}_k$, is

$$\begin{align} T &= \sum_{i=1}^{2025} \binom{R_i}{2} = \frac{1}{2} \left[ \sum_{i=1}^{2025} R_i^2 - \sum_{i=1}^{2025} R_i \right] \\ &\ge \frac{1}{2} \left[ \frac{\left(\sum_{i=1}^{2025} R_i\right)^2}{2025} - \sum_{i=1}^{2025} R_i \right] = \frac{1}{2} \left[ \frac{(90N)^2}{2025} - 90N \right] \\ &= 2N^2 - 45N, \tag{28} \end{align}$$

where in the second line we have used the Cauchy-Schwarz inequality.

Next, we give an upper bound for the number of these triples by considering pairs of matches. We know that no pair of matches has more than 3 players in common, so

$$T \le 3 \binom{N}{2} = \frac{3N(N-1)}{2}. \tag{29}$$

Combining (28) and (29) yields

$$2N^2 - 45N \leq \frac{3N(N-1)}{2}$$

which, upon simplifying, gives $N \leq 87$.

Therefore, the number of football matches is less than $90$.