15. If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive integers, and $a_{1}<a_{2}<\cdots<a_{n} \leqslant 2007$, to ensure that there always exist four distinct numbers $a_{i} 、 a_{j} 、 a_{k} 、 a_{l}$, such that $a_{i}+a_{j}=a_{k}+a_{l}=a_{n}$, what is the minimum value of $n$? And explain the reason.
15. Let $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005$, $a_{1004}=2007$, a total of 1004 odd numbers. Clearly, the sum of any two numbers is not equal to 2007.

If we add the number 2006 to the above 1004 numbers, i.e., $a_{1}=1, a_{2}=3, \cdots, a_{1003}=2005, a_{1004}=$ $2006, a_{1000}=2007$, at this point, there exists only one pair of numbers $a_{1}=1, a_{1004}=2006$, whose sum is
$$
a_{1006}=2007 \text {. }
$$

Therefore, the minimum value of $n$ is no less than 1006.
Next, we prove that when $n \geqslant 1006$, there must exist four numbers that satisfy the condition.
When $a_{n}=2007$, since
$$
2007=1+2006=2+2005=\cdots=1007+1004 \text {, }
$$

this indicates that 2007 can only be decomposed into 1003 different sums of two positive integers. Therefore, when $n \geqslant 1006$, i.e., $n-1$ $\geqslant 1005$, among the at least 1005 numbers $a_{1}, a_{2}, \cdots, a_{n-1}$ excluding $a_{n}=2007$, at least two of the 1003 different sums $a_{i}+a_{j} 、 a_{k}+a_{l}$ and their four addends $a_{i}$ 、 $a_{j} 、 a_{k} 、 a_{l}$ are included, which are the four positive integers required by the problem.

When $a_{n}<2007$, if $a_{n}=2 m-1$, then $a_{n}$ can be expressed as $m$ different sums of two different positive integers; if $a_{n}=2 m$, then $a_{n}$ can be expressed as $m-1$ different sums of two different positive integers. Since $a_{n}<2007$, we have $m \leqslant 1003$. Excluding $a_{n}$, among the at least 1005 numbers $a_{1}, a_{2}, \cdots, a_{n-1}$, at least two of the no more than 1003 different sums and their four addends are included, which are the four numbers required by the problem.
In summary, the minimum value of $n$ is 1006.