We represent each Pattano coin by an ordered pair $(i, j)$, where $i$ and $j$ are the numbers written on the blue and yellow side respectively. Let $C$ be the set of non-identical $2559$ Pattano coins. Let $b_i$ denote the number of Pattano coins in $C$ with $i$ on its blue side, and $y_j$ denote the number of Pattano coins in $C$ with $j$ on its yellow side. For a Pattano coin $(i, j)$ in $C$, let $p_{i,j}$ be the number of coins in $C$ that are pairable with $(i, j)$ (apart from itself).

It is easy to see that
$$p_{i,j} = b_i + y_j - 2,$$
thus
$$\sum_{(i,j) \in C} p_{i,j} = \sum_{(i,j) \in C} b_i + \sum_{(i,j) \in C} y_j - 5118. \quad (9)$$

Note that $b_i$ appears in the sum $\sum_{(i,j) \in C} b_i$ exactly $b_i$ times, and $y_j$ appears in the sum $\sum_{(i,j) \in C} y_j$ exactly $y_j$ times. Hence
$$\sum_{(i,j) \in C} b_i = \sum_{i=1}^{100} b_i^2 \quad \text{and} \quad \sum_{(i,j) \in C} y_j = \sum_{j=1}^{100} y_j^2. \qquad (10)$$

Applying the Cauchy-Schwarz inequality to the vectors $(b_1, b_2, \dots, b_{100})$, $(1, 1, \dots, 1)$ and $(y_1, y_2, \dots, y_{100})$, $(1, 1, \dots, 1)$ we obtain,
$$\left(\sum_{i=1}^{100} b_i\right)^2 \le \left(\sum_{i=1}^{100} 1^2\right) \left(\sum_{i=1}^{100} b_i^2\right) = 100 \sum_{i=1}^{100} b_i^2, \\ \left(\sum_{j=1}^{100} y_j\right)^2 \le \left(\sum_{j=1}^{100} 1^2\right) \left(\sum_{j=1}^{100} y_j^2\right) = 100 \sum_{j=1}^{100} y_j^2. \qquad (11)$$

Combining (9), (10) and (11) yields
$$\sum_{(i,j) \in C} p_{i,j} \ge \frac{1}{100} \left[ \left( \sum_{i=1}^{100} b_i \right)^2 + \left( \sum_{j=1}^{100} y_j \right)^2 \right] - 5118.$$

But $\sum_{i=1}^{100} b_i = \sum_{j=1}^{100} y_j = 2559$, thus
$$\sum_{(i,j) \in C} p_{i,j} \ge \frac{1}{100} (2559^2 + 2559^2) - 5118 = \frac{2559^2}{50} - 5118.$$

Finally, by the pigeonhole principle, there exists $(i,j) \in C$ such that $p_{i,j} \ge \left\lfloor \frac{2559}{50} - 2 \right\rfloor = 50$. That is, there exists a Pattano coin which is pairable with at least $50$ other coins.