14. A. Find all positive integers $n$, such that there exist positive integers $x_{1}, x_{2}, \cdots, x_{2012}$, satisfying
$$
x_{1}<x_{2}<\cdots<x_{2012} \text {, }
$$

and $\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}=n$.
14. A. Since $x_{1}, x_{2}, \cdots, x_{2012}$ are all positive integers, and $x_{1}<x_{2}<\cdots<x_{2012}$, we have
$$
x_{1} \geqslant 1, x_{2} \geqslant 2, \cdots, x_{2012} \geqslant 2012 .
$$

Then $n=\frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}$
$$
\begin{array}{l}
\leqslant \frac{1}{1}+\frac{2}{2}+\cdots+\frac{2012}{2012} \\
=2012 .
\end{array}
$$

When $n=1$, let
$$
\begin{array}{l}
x_{1}=2012, \\
x_{2}=2 \times 2012, \\
\cdots \cdots . \\
x_{2012}=2012 \times 2012 .
\end{array}
$$
$$
\text { Then } \frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}}=1 \text {. }
$$

When $n=k+1(1 \leqslant k \leqslant 2011)$, let
$$
\begin{array}{l}
x_{1}=1, x_{2}=2, \cdots, x_{k}=k, \\
x_{k+1}=(2012-k)(k+1), \\
x_{k+2}=(2012-k)(k+2), \\
x_{2012}=(2012-k) \times 2012 . \\
\text { Then } \frac{1}{x_{1}}+\frac{2}{x_{2}}+\cdots+\frac{2012}{x_{2012}} \\
=k+(2012-k) \frac{1}{2012-k} \\
=k+1=n .
\end{array}
$$

In summary, all positive integers $n$ that satisfy the condition are $1,2, \cdots, 2012$.