11. (2007 Slovakia-Slovakia-Poland Mathematics Competition) Given $n \in\{3900,3901, \cdots, 3909\}$. Find the value of $n$ such that the set $\{1,2, \cdots, n\}$ can be partitioned into several triples, and in each triple, one number is equal to the sum of the other two.
11. Since $\{1,2, \cdots, n\}$ can be partitioned into several triples, then $3 \mid n$.

Furthermore, each triple can be represented as $\{a, b, a+b\}$, so the sum of the elements in each triple is even.
Therefore, the sum of the elements in $\{1,2, \cdots, n\}$ is even, i.e., $4 \mid n(n+1)$.
In summary, $n$ must be of the form $12k$ or $12k+3$.
Thus, $n=3900$ or $n=3903$.
Finally, we need to prove that when $n=3900$ or $n=3903$, there exists a partition that satisfies the conditions.
Lemma: If there exists a partition that satisfies the conditions when $n=k$, then there also exists a partition that satisfies the conditions when $n=4k$ or $n=4k+3$.

Proof of the lemma: Since the set $\{1,2, \cdots, k\}$ satisfies the partition condition, the set $\{2,4, \cdots, 2k\}$ can also be partitioned in the same way.
For $n=4k$, the remaining numbers are
$$
\{1,3, \cdots, 2k-1,2k+1,2k+2, \cdots, 4k-1,4k\}.
$$

These can be partitioned into $k$ triples
$$
\{2j-1,3k-j+1,3k+j\}(j=1,2, \cdots, k).
$$

Each column below represents a triple:
$$
\left(\begin{array}{cccccl}
1 & 3 & 5 & \cdots & 2k-3 & 2k-1 \\
3k & 3k-1 & 3k-2 & \cdots & 2k+2 & 2k+1 \\
3k+1 & 3k+2 & 3k+3 & \cdots & 4k-1 & 4k
\end{array}\right).
$$

For $n=4k+3$, the remaining numbers are
$$
\{1,3, \cdots, 2k-1,2k+1, \cdots, 4k+2,4k+3\}.
$$

These can be partitioned into $k+1$ triples
$$
\{2j-1,3k+3-j, 3k+j+2\}(j=1,2, \cdots, k+1).
$$

Similarly, each column below represents a triple:
$$
\left(\begin{array}{cccccc}
1 & 3 & 5 & \cdots & 2k-1 & 2k+1 \\
3k+2 & 3k+1 & 3k & \cdots & 2k+3 & 2k+2 \\
3k+3 & 3k+4 & 3k+5 & \cdots & 4k+2 & 4k+3
\end{array}\right).
$$

Returning to the original problem.
By the problem statement, $n=3$ satisfies the conditions.
Therefore, by the lemma, we can obtain $n=3900$ or $n=3903$ that satisfy the partition conditions as follows:
$$
3 \xrightarrow{3 \times 4+3} 15 \xrightarrow{15 \times 4} 60 \xrightarrow{60 \times 4+3} 243 \xrightarrow{243 \times 4+3} 975{ }_{3903}^{3900}
$$