Given the relation $(x_1, y_1) R (x_2, y_2)$ if $x_1 \le x_2$ or $y_1 \le y_2$.
For reflexivity, we need to check if $(x_1, y_1) R (x_1, y_1)$.
Since $x_1 \le x_1$ and $y_1 \le y_1$, the condition $x_1 \le x_1$ or $y_1 \le y_1$ is true. Therefore, $R$ is reflexive.
For symmetry, if $(x_1, y_1) R (x_2, y_2)$, then $x_1 \le x_2$ or $y_1 \le y_2$. For $R$ to be symmetric, $(x_2, y_2) R (x_1, y_1)$ must also be true, which means $x_2 \le x_1$ or $y_2 \le y_1$. This is not necessarily true. For example, $(1, 2) R (3, 4)$ since $1 \le 3$ or $2 \le 4$, but $(3, 4) R (1, 2)$ is false because neither $3 \le 1$ nor $4 \le 2$ is true. Thus, $R$ is not symmetric.
For transitivity, consider pairs $(3, 9)$, $(4, 6)$, and $(2, 7)$.
$(3, 9) R (4, 6)$ because $3 \le 4$.
$(4, 6) R (2, 7)$ because $6 \le 7$.
However, $(3, 9) R (2, 7)$ is false because neither $3 \le 2$ nor $9 \le 7$ is true. Therefore, $R$ is not transitive.
Since $R$ is reflexive but not symmetric, statement (I) is correct.
Since $R$ is not transitive, statement (II) is incorrect.
Thus, only statement (I) is correct.