14. An extracurricular activity group at a school designed a tree planting plan on graph paper for a certain desert as follows: The $k$-th tree is planted at point $P_{k}\left(x_{k}, y_{k}\right)$, where $x_{1}=1, y_{1}=1$. When $k \geqslant 2$,
$$
\left\{\begin{array}{l}
x_{k}=x_{k-1}+1-5\left[\frac{k-1}{5}\right]+5\left[\frac{k-2}{5}\right], \\
y_{k}=y_{k-1}+\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right],
\end{array}\right.
$$

where $[a]$ denotes the integer part of the real number $a$ (for example, $[2.6]=2, [0.6]=0$). According to this plan, the coordinates of the point where the 2008th tree is planted are
14. $(3,402)$.

Let $f(k)=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]$. Then
$$
\begin{array}{l}
f(k+5)=\left[\frac{k+5-1}{5}\right]-\left[\frac{k+5-2}{5}\right] \\
=\left[1+\frac{k-1}{5}\right]-\left[1+\frac{k-2}{5}\right] \\
=\left[\frac{k-1}{5}\right]-\left[\frac{k-2}{5}\right]=f(k) .
\end{array}
$$

Therefore, $f(k)$ is a periodic function with a period of 5.
Calculations show that
$$
\begin{array}{l}
f(2)=0, f(3)=0, f(4)=0, \\
f(5)=0, f(6)=1 .
\end{array}
$$

Thus, $x_{2008}=x_{2007}+1-5 f(2008)$,
$$
x_{2007}=x_{2000}+1-5 f(2007), \cdots \ldots .
$$
$$
x_{2}=x_{1}+1-5 f(2) \text {. }
$$

Adding all the above equations, we get
$$
\begin{array}{l}
x_{2008}=x_{1}+2007-5[f(2)+f(3)+\cdots+f(2008)] \\
=x_{1}+2007-5[401(f(2)+f(3)+\cdots+ \\
f(6))+f(2)+f(3)] \\
=x_{1}+2007-5 \times 401=3 .
\end{array}
$$

Similarly, $y_{2008}=402$.
Therefore, the planting point of the 2008th tree is $(3,402)$.