Find all triples of positive integers $(a, b, c)$ satisfying $a^{2}+b^{2}+c^{2}=2005$, and $a \leqslant b \leqslant c$.
Since any odd square number divided by 4 leaves a remainder of 1, and any even square number is a multiple of 4, and 2005 divided by 4 leaves a remainder of 1, it follows that among $a^{2}$, $b^{2}$, and $c^{2}$, there must be two even square numbers and one odd square number.

Let $a=2m$, $b=2n$, $c=2k-1$, where $m$, $n$, and $k$ are positive integers (for the sake of calculation, we temporarily do not consider the size relationship between $a$, $b$, and $c$). Then the original equation can be transformed into
$$
m^{2}+n^{2}+k(k-1)=501 \text {. }
$$

Since any square number divided by 3 leaves a remainder of either 0 or 1, we now discuss $k$:
(1) If $3 \mid k(k-1)$, then from equation (1) we know $3 \mid \left(m^{2}+n^{2}\right)$. Therefore, $m$ and $n$ are both multiples of 3. Let $m=3m_{1}$, $n=3n_{1}$, and $\frac{k(k-1)}{3}$ is an integer. From equation (1) we have
$$
3 m_{1}^{2}+3 n_{1}^{2}+\frac{k(k-1)}{3}=167 \text {. }
$$
Thus, $\frac{k(k-1)}{3} \equiv 167=2(\bmod 3)$.
Let $\frac{k(k-1)}{3}=3r+2$. Then
$$
k(k-1)=9r+6 \text {. }
$$
Also, from equation (1) we have $k(k-1)<501$, so $k \leqslant 22$.
Therefore, from equation (3), $k$ can take the values $3, 7, 12, 16, 21$, and substituting them into equation (2) we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
k=3, \\
m_{1}^{2}+n_{1}^{2}=55;
\end{array}\right. \\
\left\{\begin{array}{l}
k=7, \\
m_{1}^{2}+n_{1}^{2}=51;
\end{array}\right. \\
\left\{\begin{array}{l}
k=12, \\
m_{1}^{2}+n_{1}^{2}=41;
\end{array}\right. \\
\left\{\begin{array}{l}
k=16, \\
m_{1}^{2}+n_{1}^{2}=29;
\end{array}\right. \\
\left\{\begin{array}{l}
k=21, \\
m_{1}^{2}+n_{1}^{2}=9.
\end{array}\right.
\end{array}
$$

Since $55$ and $51$ are both of the form $4N+3$, they cannot be expressed as the sum of two squares, and 9 cannot be expressed as the sum of two positive integers' squares, therefore, only when $k=12$ and $k=16$, there are positive integer solutions for $m_{1}$ and $n_{1}$.
When $k=12$, we get $\left(m_{1}, n_{1}\right)=(4,5)$, then
$$
a=6m_{1}=24, \quad b=6n_{1}=30, \quad c=2k-1=23 \text {. }
$$

Therefore, $(a, b, c)=(24,30,23)$.
When $k=16$, we get $\left(m_{1}, n_{1}\right)=(2,5)$, then
$$
a=6m_{1}=12, \quad b=6n_{1}=30, \quad c=2k-1=31 \text {. }
$$

Therefore, $(a, b, c)=(12,30,31)$.
(2) If $3 \nmid k(k-1)$, since any three consecutive numbers must include one multiple of 3, then $k+1$ is a multiple of 3. Hence, $k$ leaves a remainder of 2 when divided by 3. Therefore, $k$ can only take the values $2, 5, 8, 11, 14, 17, 20$.
Using equation (1) to discuss each case as follows.
If $k=2$, then $m^{2}+n^{2}=499$. Since $499 \equiv 3(\bmod 4)$, there is no solution in this case.
If $k=5$, then $m^{2}+n^{2}=481$. Using the identity
$$
\begin{array}{l}
\left(\alpha^{2}+\beta^{2}\right)\left(x^{2}+y^{2}\right)=(\alpha x+\beta y)^{2}+(\alpha y-\beta x)^{2} \\
=(\alpha x-\beta y)^{2}+(\alpha y-\beta x)^{2}, \\
481=13 \times 37=\left(3^{2}+2^{2}\right)\left(6^{2}+1^{2}\right) \\
=20^{2}+9^{2}=15^{2}+16^{2} \text {. }
\end{array}
$$

We know $481=13 \times 37=\left(3^{2}+2^{2}\right)\left(6^{2}+1^{2}\right)$
So, $(m, n)=(20,9)$ or $(15,16)$.
Thus, $(a, b, c)=(2m, 2n, 2k-1)=(40,18,9)$ or $(30,32,9)$.

If $k=8$, then $m^{2}+n^{2}=445$. Since $445=5 \times 89=\left(2^{2}+1^{2}\right)\left(8^{2}+5^{2}\right)=21^{2}+2^{2}=18^{2}+11^{2}$,
So, $(m, n)=(21,2)$ or $(18,11)$.
Thus, $(a, b, c)=(2m, 2n, 2k-1)=(42,4,15)$ or $(36,22,15)$.

If $k=11$, then $m^{2}+n^{2}=391$. Since $391 \equiv 3(\bmod 4)$, there is no solution in this case.

If $k=14$, then $m^{2}+n^{2}=319$. Since $319 \equiv 3(\bmod 4)$, there is no solution in this case.

If $k=17$, then $m^{2}+n^{2}=229$. Since $229=15^{2}+2^{2}$, so, $(m, n)=(15,2)$. Thus,
$$
(a, b, c)=(2m, 2n, 2k-1)=(30,4,33) \text {. }
$$

If $k=20$, then $m^{2}+n^{2}=121=11^{2}$. Since $11^{2}$ cannot be expressed as the sum of two positive integers' squares, there is no solution in this case.
Therefore, there are 7 solutions that satisfy $a \leqslant b \leqslant c$:
$$
\begin{array}{l}
(23,24,30),(12,30,31),(9,18,40),(9,30,32), \\
(4,15,42),(15,22,36),(4,30,33) \\
\text { Upon verification, they all satisfy the equation. }
\end{array}
$$

Upon verification, they all satisfy the equation.