Parse the largest $n$ is 561.
With lines parallel to the sides of the square matrix as the $x$-axis and $y$-axis. Take the two points in the square with the smallest distance as the unit length.

Let the convex $n$-gon $T$ satisfy the conditions. For each side $p$ of the convex polygon $T$, if the projections of $p$ on the $x$-axis and $y$-axis are $x_{p}$ and $y_{p} (x_{p}, y_{p} \in \mathbf{N})$, we call this side a $\left(\frac{x_{p}}{\left(x_{p}, y_{p}\right)}, \frac{y_{p}}{\left(x_{p}, y_{p}\right)}\right)$-shaped side. Clearly, the slope of an $(a, b)$-shaped side is $\pm \frac{b}{a}. ((0,1)$-shaped side has a slope of $\infty)$
Since $T$ is a convex polygon, there are at most 2 sides with a slope of $\frac{b}{a}$, and at most 2 sides with a slope of $-\frac{b}{a}$. Therefore, when $a, b \in \mathbf{N}^{*}, (a, b)=1$, there are at most 4 $(a, b)$-shaped sides; $(1,0)$-shaped sides and $(0,1)$-shaped sides each have at most 2 sides.
Consider the total length of the projections of all sides of $T$ on the $x$-axis and $y$-axis.
Since each line parallel to the $y$-axis intersects with at most two sides of $T$, each point on the $x$-axis is on the projection of at most two sides of $T$ (vertices of $T$ are not considered as points on the sides of $T$). Therefore, the total projection length of $T$ on the $x$-axis does not exceed
$$
\sum_{p \text { is a side of } T}\left(x_{p}+y_{p}\right) \leqslant 4 \cdot 2003.
$$

For an $(a, b)$-shaped side $p$, we call $a+b$ the side length of $p$. Clearly,
$$
x_{p}+y_{p} \geqslant a+b.
$$
The side length of each side of $T$ is a positive integer. We consider how many sides have a side length of $k$.
For each $k \in \mathbf{N}^{*}$, we have
$$
S_{k}=\left\{(a, b) \mid a, b \in \mathbf{N}^{*}, (a, b)=1, a+b=k\right\},
$$

Clearly, $S_{1}=\{(0,1),(1,0)\}$.
For $k \geqslant 2$, we prove that
$$
\left|S_{k}\right|=\varphi(k) (\varphi(k) \text{ is the Euler's totient function). }
$$

Since $(a, b)=1, a+b=k$, we have
$$
(a, a+b)=(a, k)=1, 1 \leqslant a \leqslant k-1,
$$

Therefore, $a$ can only take the numbers that are coprime with $k$ from $1, 2, \cdots, k$, and there are $\varphi(k)$ such numbers.
Conversely, as long as $(a, k)=1, 1 \leqslant a \leqslant k-1$, let $b=k-a$, then $(a, b)=1$. Therefore, $(a, b) \in S_{k}$.
So $\left|S_{k}\right|=\varphi(k)$.
Let $T$ have $a_{k}$ sides with side length $k$ $(k=1,2, \cdots)$, then $a_{1}+\cdots+a_{k}+\cdots=n$. From (4) and (1), we know
$$
a_{1} \leqslant 4, a_{k} \leqslant 4 \varphi(k),
$$

So from (3) we get
The table below lists the values of $\varphi(k)$ and $k \varphi(k)$ for $k=2,3, \cdots, 22$:
Notice that
$$
\sum_{k=2}^{21} \varphi(k)=139, \sum_{k=2}^{21} k \varphi(k)=1996,
$$

When $n \geqslant 562$, from (5) we know
$$
\begin{aligned}
\sum_{k=1} k a_{k} & \geqslant 1 \cdot 4+4 \sum_{k=2}^{21} k \varphi(k)+22\left(n-4 \sum_{k=2}^{21} \varphi(k)-4\right) \\
& \geqslant 1 \cdot 4+4 \cdot 1996+22 \cdot(562-560)=8032 .
\end{aligned}
$$

This contradicts (2) and (6)! Therefore, $n \leqslant 561$.
We will provide an example for $n=561$.
Consider the set of 561 vectors (denoted as $S_{0}$);
$$
\begin{aligned}
S_{0}= & \left\{( \pm a, \pm b) \mid(a, b) \in S_{2} \cup S_{3} \cup \cdots\right. \\
& \left.\cup S_{21} \triangleq S\right\} \cup\{( \pm 1,0),(0, \pm 1), \\
& (3,19),(3,-19),(-15,7)\} \\
& -\{(-8,-13),(-1,20)\} .
\end{aligned}
$$
( $\left|S_{0}\right|=4 \cdot 139+7-2=561$ )
We will prove:
(1) The vectors in $S_{0}$ are not collinear;
(2) The sum of the vectors in $S_{0}$ is 0;
(3) The sum of the absolute values of the $x$-coordinates of the vectors in $S_{0}$ is $2 \cdot 2003$, and the sum of the absolute values of the $y$-coordinates of the vectors in $S_{0}$ is $2 \cdot 2003$.
(1) Suppose $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) \in S_{0},\left(x_{1}, y_{1}\right)$ is collinear with $\left(x_{2}, y_{2}\right)$, i.e., $\frac{x_{2}}{x_{1}}=$ $\frac{y_{2}}{y_{1}} \Rightarrow x_{2} y_{1}=x_{1} y_{2}$.

Since $\left(x_{1}, y_{1}\right)=1,\left(x_{2}, y_{2}\right)=1$, and $x_{2} \mid x_{1} y_{2}$, we have $x_{2} \mid x_{1}$, and similarly $x_{1} \mid x_{2}$, thus $x_{2}= \pm \dot{x}_{1}$.
Since $\left(x_{1}, y_{1}\right)$ is collinear with $\left(x_{2}, y_{2}\right)$, then $x_{1} 、 x_{2}$ have the same sign, so $x_{2}=x_{1}$. Similarly, $y_{2}=y_{1}$.
(2) $\sum_{a \in S_{0}} \alpha=\sum_{(a, b) \in s}[(a, b)+(a,-b)+(-a, b)+(-a,-b)]+[(1,0)+$ $(0,1)+(-1,0)+(0,-1)+(3,19)+(3,-19)+(-15,7)]-[(-8,-13)$ $+(-1,20)]=0$. $+(-1,20)]=0$.
(3) $\sum_{a \in S_{0}}\left|x_{a_{0}}\right|=\left(\sum_{(a, b) \in S} 4 a\right)+(1+1+3+3+15-8-1)=2 \sum_{(a, b) \in s}(a+b)+$ $14=2 \sum_{k=2}^{21}\left|S_{k}\right| k+14=3992+14=2 \cdot 2003$.
(Since $(a, b) \in S$, then $(b, a) \in S$, so $\sum_{(a, b) \in S} 4 a=\sum_{(a, b) \in S} 4 b=2 \sum_{(a, b) \in S}(a+b)$ )
Similarly, $\sum_{0 \in S_{0}}\left|y_{\alpha_{0}}\right|=2 \cdot 2003$.
Next, we construct a convex 561-gon using the vectors in $S_{0}$.
As shown in Figure a, arrange the vectors in $S_{0}$ in counterclockwise order as $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{a}_{561}$ (by (1)). Let $A_{0}=$ origin $O$, find $A_{k}$ such that $\overrightarrow{A_{0} A_{k}}=\boldsymbol{\alpha}_{1}+\cdots+\boldsymbol{\alpha}_{k}(k=1,2, \cdots, 561)$, then $\overrightarrow{A_{0} A_{561}}=$ $\boldsymbol{a}_{1}+\cdots+\boldsymbol{a}_{561}=0$ (by (2)). Therefore, $A_{561}$ coincides with $A_{0}$. This shows that $A_{0} A_{1} \cdots A_{560}$ forms a 561-gon, denoted as $T$.
Since $\alpha_{1}, \cdots, \alpha_{561}$ are arranged counterclockwise, $T$ is convex. (As shown in Figure b)
Since the coordinates of $a_{i}$ are all integers, $T$ is a lattice polygon.
Figure a
By (3), the projections of $T$ on the $x$-axis and $y$-axis are both 2003.
24. 3. 15 On a plane, there are $2004 \times 2004$ points arranged in a square array. Find the largest $n$ such that a convex $n$-gon can be drawn, with all its vertices being points from the array.