L, 'Positive integer answers form an arithmetic sequence, the product of its first and fourth numbers equals the larger root of the equation $x^{1+\lg x}=(0.001)^{-x^{2}}$, and the sum of the squares of the second and third numbers is five times the number of terms in the expansion of $\left[\frac{1}{4 \sqrt{x}}+x^{-\frac{1}{6}}\right]^{20}$ that do not contain $\mathrm{x}$. Find these four numbers.
Seven. Solution: $\because \mathrm{x}^{1+1 g \mathrm{~g}}=(0.001)^{-\frac{2}{3}}$.
That is, $x^{1}+\lg x=1000^{\frac{2}{3}}$.
Taking the logarithm, we get $(1+\lg x) \lg x=\frac{2}{3} \lg 1000$,
$$
\begin{array}{c}
\lg ^{2} x+\lg x-2=0, \\
\lg x=-2, \lg x=1, \\
\therefore \quad x_{1}=\frac{1}{100}, \quad x_{2}=10 .
\end{array}
$$

Upon verification, $x_{1}$ and $x_{2}$ are both roots of the original equation. According to the problem, we take $x=10$. Review,
$$
\begin{aligned}
\mathbf{T}_{k+1} & =\mathbf{C}_{20}^{k} \cdot\left(x^{6}\right)^{\frac{1}{4}} \cdot\left(\frac{1}{4 /-x^{2}}\right)^{20-k} \\
& =\mathbf{C}_{20}^{k} \mathbf{x}^{\frac{k}{6}}-\left(\frac{20-k}{4}\right) \\
& =\mathbf{C}_{20}^{k} \cdot x^{\frac{5 k-60}{12}} . \\
\because \quad & \frac{5 k-60}{12}=0, \therefore 5 k-60=0, \\
\therefore k & =12 .
\end{aligned}
$$

Therefore, the number of terms without $x$ is $k+1=13$.
Let the four numbers be $a-3 y, a-y$,
$$
a+y, a+3 y \text {. }
$$

According to the problem, we get
$$
\left\{\begin{array}{l}
(a-y)^{2}+(a+y)^{2}=13 \times 5, \\
(a-3 y)(a+3 y)=10
\end{array}\right.
$$

Solving the system of equations, we get
$$
\mathrm{y}_{1}=\frac{3}{2}, \mathrm{y}_{2}=-\frac{3}{2} \text { (not in line with the problem, discard). }
$$

Substituting $\mathrm{y}=\frac{3}{2}$ into (2), we get
$\mathrm{a}= \pm \frac{11}{2}, \quad \mathrm{a}=-\frac{11}{2}$ (not in line with the problem),
$$
\begin{aligned}
\quad \therefore \quad a & =\frac{11}{2} . \\
\therefore \quad a-3 y & =1, a-y=4, a+y=7, \\
a+3 y & =10 .
\end{aligned}
$$

Therefore, the four positive numbers are $1,4,7,10$.