Let $R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : \log_e(x + y) \leq 2\}$. Then the minimum number of elements, required to be added in $R$ to make it a transitive relation, is __________.

Given the relation $R = \{(x, y) \in \mathbb{N} \times \mathbb{N} : \log_e(x + y) \leq 2\}$.

This condition simplifies to:
$x + y \leq e^2$

Since $e \approx 2.718$, we have $e^2 \approx 7.389$.
As $x, y \in \mathbb{N}$, the sum $x + y$ must be a natural number. Therefore, the condition becomes:
$x + y \leq 7$

Let's find the number of elements in $R$ by listing the possible pairs $(x, y)$:
For $x = 1$, $y \in \{1, 2, 3, 4, 5, 6\}$ (6 elements)
For $x = 2$, $y \in \{1, 2, 3, 4, 5\}$ (5 elements)
For $x = 3$, $y \in \{1, 2, 3, 4\}$ (4 elements)
For $x = 4$, $y \in \{1, 2, 3\}$ (3 elements)
For $x = 5$, $y \in \{1, 2\}$ (2 elements)
For $x = 6$, $y \in \{1\}$ (1 element)

Total number of elements in $R = 6 + 5 + 4 + 3 + 2 + 1 = 21$.

To make $R$ a transitive relation, we must ensure that if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
Notice that $(x, 1) \in R$ for all $x \in \{1, 2, 3, 4, 5, 6\}$ and $(1, z) \in R$ for all $z \in \{1, 2, 3, 4, 5, 6\}$.
For the relation to be transitive, it must contain $(x, z)$ for all possible combinations of $x$ and $z$ from the set $A = \{1, 2, 3, 4, 5, 6\}$.

Thus, the transitive closure of $R$ will be the Cartesian product $A \times A$, which contains $6 \times 6 = 36$ elements.

The minimum number of elements required to be added to $R$ to make it transitive is the difference between the number of elements in the transitive closure and the number of elements already in $R$:
Minimum elements to add $= 36 - 21 = 15$.

Answer: $15$