Problem 1340

1. Understanding the Problem:
We need to determine how many positive integers $n$ in the range $1 \leq n \leq 2008$ allow the set $\{1, 2, 3, \ldots, 4n\}$ to be divided into $n$ disjoint 4-element subsets such that each subset contains an element which is the arithmetic mean of all the elements in that subset.

2. Arithmetic Mean Condition:
If $a, b, c, d$ are four numbers in a subset and $a$ is the arithmetic mean, then:
$$\frac{a + b + c + d}{4} = a \implies a + b + c + d = 4a$$
This implies that the sum $a + b + c + d$ must be divisible by 4.

3. Sum of the Entire Set:
The sum of all elements in the set $\{1, 2, 3, \ldots, 4n\}$ is:
$$\sum_{k=1}^{4n} k = \frac{4n(4n + 1)}{2} = 2n(4n + 1)$$
For the set to be divided into $n$ subsets each with a sum divisible by 4, the total sum $2n(4n + 1)$ must be divisible by 4.

4. Divisibility Condition:
$$2n(4n + 1) \equiv 0 \pmod{4}$$
Simplifying, we get:
$$2n(4n + 1) \equiv 0 \pmod{4} \implies n(4n + 1) \equiv 0 \pmod{2}$$
Since $4n + 1$ is always odd, $n$ must be even for the product $n(4n + 1)$ to be even.

5. **Counting Valid $n$:**
We need to count the even integers $n$ in the range $1 \leq n \leq 2008$. The even numbers in this range are $2, 4, 6, \ldots, 2008$. This is an arithmetic sequence with the first term $a = 2$ and common difference $d = 2$.

6. Number of Terms in the Sequence:
The number of terms $k$ in the sequence can be found using:
$$a_k = a + (k-1)d \implies 2008 = 2 + (k-1) \cdot 2 \implies 2008 = 2k \implies k = 1004$$

Conclusion:
The number of positive integers $n$ such that $1 \leq n \leq 2008$ and the set $\{1, 2, 3, \ldots, 4n\}$ can be divided into $n$ disjoint 4-element subsets each containing its arithmetic mean is $\boxed{1004}$.