14. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=a_{2}=1, a_{3}=m, \\
a_{n+1}=\frac{k+a_{n} a_{n-1}}{a_{n-2}}(n \geqslant 3),
\end{array}
$$

where $k, m \in \mathbf{Z}_{+}$, and $(k, m)=1$. For what value of $k$ will $a_{n}$ be an integer for all $n \in \mathbf{Z}_{+}$?
14. From the problem, we know
$$
\begin{array}{l}
a_{4}=k+m, a_{5}=k+m(k+m), \\
a_{6}=\frac{(k+m)\left(m^{2}+k m+k\right)+k}{m} .
\end{array}
$$

On one hand, from
$$
\begin{array}{l}
(k+m)\left(m^{2}+k m+k\right)+k \\
=k(k+1)+m\left(m^{2}+2 k m+k^{2}+k\right),
\end{array}
$$

and $(k, m)=1$, thus, if $a_{6}$ is an integer, then $m \mid(k+1)$.
Therefore, we can set $k=t m-1(t \in \mathbb{Z})$.
On the other hand, let $k=t m-1(t \in \mathbb{Z})$. Then
$$
\begin{array}{l}
a_{n+1} a_{n-2}=k+a_{n} a_{n-1}, \\
a_{n+2} a_{n-1}=k+a_{n+1} a_{n} .
\end{array}
$$

Thus, $a_{n+1}\left(a_{n-2}+a_{n}\right)=a_{n-1}\left(a_{n}+a_{n+2}\right)$
$$
\Rightarrow \frac{a_{n+2}+a_{n}}{a_{n+1}}=\frac{a_{n}+a_{n-2}}{a_{n-1}} \text {. }
$$

Assume that when $n=s$, $a_{1}, a_{2}, \cdots, a_{s}$ are integers.
If $s+1$ is odd, then
$$
\begin{array}{l}
\frac{a_{s+1}+a_{s-1}}{a_{s}}=\frac{a_{s-1}+a_{s-3}}{a_{s-2}} \\
=\cdots=\frac{a_{1}+a_{3}}{a_{2}}=m+1 .
\end{array}
$$

Thus, $a_{s+1}=(m+1) a_{s}-a_{s-1}$ is an integer.
If $s+1$ is even, then
$$
\begin{array}{l}
\frac{a_{s+1}+a_{s-1}}{a_{s}}=\frac{a_{s-1}+a_{s-3}}{a_{s-2}} \\
=\cdots=\frac{a_{4}+a_{2}}{a_{3}}=t+1 .
\end{array}
$$

Thus, $a_{s+1}=(t+1) a_{s}-a_{s-1}$ is an integer.
In summary, when $k=t m-1(t \in \mathbb{Z})$, $a_{n}$ is an integer.