10. Given that there are exactly 600 triangles with integer sides, all of different lengths, and the longest side is exactly $n$. Find $n$.

The longest side of the triangle is $n$, and the other two sides are $a$ and $b$ with $a < b < n$. The triangle inequality theorem states that:
1. $a + b > n$
2. $a + n > b$
3. $b + n > a$

Since $a < b < n$, the second and third inequalities are always satisfied. Therefore, the only relevant inequality is $a + b > n$.

To find the number of such triangles, we need to count the number of pairs $(a, b)$ that satisfy $a < b < n$ and $a + b > n$.

For a fixed $b$, $a$ can take values from $b - n + 1$ to $b - 1$. The number of valid $a$ values for a given $b$ is $b - \left\lceil \frac{n + 1}{2} \right\rceil + 1$.

Summing over all possible values of $b$ from $\left\lceil \frac{n + 1}{2} \right\rceil$ to $n - 1$, we get the total number of triangles:
\[
\sum_{b=\left\lceil \frac{n + 1}{2} \right\rceil}^{n-1} (b - \left\lceil \frac{n + 1}{2} \right\rceil + 1)
\]

This sum can be simplified to:
\[
\sum_{k=1}^{n - \left\lceil \frac{n + 1}{2} \right\rceil} k = \frac{(n - \left\lceil \frac{n + 1}{2} \right\rceil)(n - \left\lceil \frac{n + 1}{2} \right\rceil + 1)}{2}
\]

Given that this sum equals 600, we can solve for $n$:
\[
\frac{(n - \left\lceil \frac{n + 1}{2} \right\rceil)(n - \left\lceil \frac{n + 1}{2} \right\rceil + 1)}{2} = 600
\]

By trial and error or solving the quadratic equation, we find that $n = 49$ satisfies this equation. Therefore, the value of $n$ is:
\[
\boxed{49}
\]
Analysis: Since one side of the triangle is fixed at length $n$, and the three side lengths $x, y, n$ are all integers, the number of such triangular shapes corresponds one-to-one with the lattice points $(x, y)$ in the coordinate plane. Therefore, the lattice point method can be used to solve this problem.

Let the lengths of the other two sides of the triangle be $x, y$. Thus, the number of positive integer solutions satisfying $x+y>n, x<y<n$ is the number of triangles.

When $n$ is odd, the number of triangles is the number of lattice points within $\triangle B C D$ in Figure 1. That is,
$$
\begin{array}{l}
2+4+6+\cdots+(n-3) \\
=\frac{1}{4} n(n-3) .
\end{array}
$$

When $n$ is even, the number of triangles is the number of lattice points within $\triangle B C D$ in Figure 2, which is
$1+3+5+\cdots+(n-3)=\frac{1}{4}(n-2)^{2}$.
When $-\frac{1}{4} n(n-3)=600$, solving gives $n=51$:
When $\frac{1}{4}(n-2)^{2}=600$, there is no integer solution.
Therefore, the required $n=51$.