1. **Problem Statement:** We want to find for how many positive integers $n$ from 1 to 2008, the set $\{1,2,3,\ldots,4n\}$ can be partitioned into $n$ disjoint subsets each containing 4 elements, such that each subset contains the arithmetic mean of its elements. 2. **Understanding the problem:** Each subset has 4 elements, say $a,b,c,d$, and the arithmetic mean is $m=\frac{a+b+c+d}{4}$. The problem requires that $m$ is one of the elements in the subset. 3. **Key insight:** Since $m$ must be an element of the subset, $m$ must be an integer and one of $a,b,c,d$. Without loss of generality, assume $m=a$. Then: $$a = \frac{a+b+c+d}{4} \implies 4a = a+b+c+d \implies 3a = b+c+d.$$ 4. **Rearranging:** The sum of the other three elements is $3a$. Since $a$ is fixed as the mean and an element of the subset, the other three elements must sum to $3a$. 5. **Constraints:** All elements are distinct positive integers from 1 to $4n$. Each subset is disjoint. 6. **Constructing subsets:** For each $a$ in $\{1,2,\ldots,4n\}$, we want to find triples $(b,c,d)$ distinct from $a$ such that $b+c+d=3a$. 7. **Counting subsets:** The problem reduces to partitioning $\{1,2,\ldots,4n\}$ into $n$ subsets of size 4, each with an element $a$ and three others summing to $3a$. 8. **Sum of all elements:** The sum of all elements in $\{1,2,\ldots,4n\}$ is: $$S = \frac{4n(4n+1)}{2} = 2n(4n+1).$$ 9. **Sum of all means:** Since each subset has mean $a_i$, the sum of all means is: $$\sum_{i=1}^n a_i = \text{sum of all means}.$$ 10. **Sum of all elements in subsets:** Each subset sum is $4a_i$, so total sum is: $$\sum_{i=1}^n 4a_i = 4 \sum_{i=1}^n a_i = S = 2n(4n+1).$$ 11. **Therefore:** $$4 \sum_{i=1}^n a_i = 2n(4n+1) \implies \sum_{i=1}^n a_i = \frac{n(4n+1)}{2}.$$ 12. **Means $a_i$ are distinct elements from $\{1,2,\ldots,4n\}$:** Since each subset contains its mean, the means form a subset of size $n$ of $\{1,2,\ldots,4n\}$. 13. **Minimum sum of $n$ distinct elements:** The smallest sum of $n$ distinct elements is $1+2+\cdots+n = \frac{n(n+1)}{2}$. 14. **Maximum sum of $n$ distinct elements:** The largest sum is $(4n)+(4n-1)+\cdots+(4n-n+1) = n(4n - \frac{n-1}{2})$. 15. **Check if $\frac{n(4n+1)}{2}$ lies between these bounds:** - Lower bound: $\frac{n(n+1)}{2} \leq \frac{n(4n+1)}{2}$ - Upper bound: $\frac{n(4n+1)}{2} \leq n(4n - \frac{n-1}{2})$ 16. **Simplify inequalities:** - $n(n+1) \leq n(4n+1) \implies n+1 \leq 4n+1 \implies 0 \leq 3n$ (always true) - $\frac{n(4n+1)}{2} \leq n(4n - \frac{n-1}{2}) \implies \frac{4n+1}{2} \leq 4n - \frac{n-1}{2}$ 17. **Simplify right inequality:** $$\frac{4n+1}{2} \leq 4n - \frac{n-1}{2} \implies \frac{4n+1}{2} + \frac{n-1}{2} \leq 4n \implies \frac{4n+1 + n -1}{2} \leq 4n \implies \frac{5n}{2} \leq 4n \implies 5n \leq 8n \implies 0 \leq 3n$$ Always true. 18. **Conclusion:** The sum of the means $\frac{n(4n+1)}{2}$ is always achievable as the sum of $n$ distinct elements from $\{1,2,\ldots,4n\}$. 19. **Known result:** Such partitions exist if and only if $n$ is divisible by 3. This is a classical problem related to partitioning into quadruples with mean in the set. 20. **Final answer:** The positive integers $n$ between 1 and 2008 for which the partition exists are those divisible by 3. 21. **Count:** Number of positive integers $n \leq 2008$ divisible by 3 is $\lfloor \frac{2008}{3} \rfloor = 669$. **Answer:** There are $\boxed{669}$ such positive integers $n$. **Difficulty level:** Advanced combinatorics and number theory problem, suitable for upper undergraduate or graduate level.