Out of `25` consecutive natural numbers, two numbers can be chosen in `lambda` ways such that their "sum" is divisible by `2`. Then `lambda//36` is.

Explanation

To solve the problem, we need to find the number of ways to choose two numbers from 25 consecutive natural numbers such that their sum is divisible by 2. This can happen in two scenarios: both numbers are even or both numbers are odd.

### Step-by-Step Solution:

1. **Identify the Set of Numbers**:
Let's consider the 25 consecutive natural numbers starting from 1 to 25:
\[ 1, 2, 3, \ldots, 25 \]

2. **Count Even and Odd Numbers**:
In this range, the even numbers are:
\[ 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 \quad (\text{Total: } 12 \text{ even numbers}) \]
The odd numbers are:
\[ 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25 \quad (\text{Total: } 13 \text{ odd numbers}) \]

3. **Choose Two Even Numbers**:
The number of ways to choose 2 even numbers from 12 is given by the combination formula:
\[ \binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66 \]

4. **Choose Two Odd Numbers**:
The number of ways to choose 2 odd numbers from 13 is:
\[ \binom{13}{2} = \frac{13!}{2!(13-2)!} = \frac{13 \times 12}{2 \times 1} = 78 \]

5. **Calculate Total Ways (λ)**:
The total number of ways to choose two numbers such that their sum is divisible by 2 (either both even or both odd) is:
\[ \lambda = \binom{12}{2} + \binom{13}{2} = 66 + 78 = 144 \]

6. **Calculate λ/36**:
Now, we need to find \( \frac{\lambda}{36} \):
\[ \frac{144}{36} = 4 \]

### Final Answer:
\[ \lambda // 36 = 4 \]