258. It is known that

$$
\begin{array}{r}
a_{1}-4 a_{2}+3 a_{3} \geqslant 0 \\
a_{2}-4 a_{3}+3 a_{4} \geqslant 0 \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
a_{98}-4 a_{99}+3 a_{100} \geqslant 0 \\
a_{99}-4 a_{100}+3 a_{1} \geqslant 0 \\
a_{100}-4 a_{1}+3 a_{2} \geqslant 0
\end{array}
$$

Let $a_{1}=1$; what are the numbers $a_{2}, a_{3}, \ldots, a_{100}$ then?
258. Adding the left parts of all our inequalities, we get the sum of the numbers $a_{1}, \alpha_{2}, \ldots, a_{99}, a_{100}$, each taken with the coefficient $1+(-4)+3=0$, i.e., the number 0. But if the sum of 100 non-negative numbers is zero, then all these numbers are zero; thus, our system of inequalities actually turns into a system of equalities:

$$
a_{1}-4 a_{2}+3 a_{3}=0, \alpha_{2}-1 a_{3}+3 a_{4}=0, \ldots, a_{160}-4 a_{1}+3 a_{2}=0
$$

which can also be rewritten as follows:

$a_{1}-a_{2}=3\left(a_{2}-a_{3}\right), a_{2}-a_{3}=3\left(a_{3}-a_{4}\right), a_{3}-a_{4}=$
$=3\left(a_{4}-a_{5}\right), \ldots, \quad a_{99}-a_{100}=3\left(a_{100}-a_{1}\right), \quad a_{100}-a_{1}=3\left(a_{1}-a_{2}\right)$.

Now we sequentially obtain:

$a_{1}-a_{2}=3\left(a_{2}-a_{3}\right)=3^{2}\left(a_{3}-a_{4}\right)=3^{3}\left(a_{4}-a_{5}\right)=\ldots$

$$
\ldots=3^{99}\left(a_{100}-a_{1}\right)=3^{100}\left(a_{1}-a_{2}\right)
$$

But from the equality $a_{1}-a_{2}=3^{100}\left(a_{1}-a_{2}\right)$ it follows that $a_{1}-a_{2}=0$; therefore, $a_{2}-a_{3}=\frac{1}{3}\left(a_{1}-a_{2}\right)=0$, and $a_{3}-a_{4}=\frac{1}{3}\left(a_{2}-a_{3}\right)=$ $=0, \ldots$, and $a_{100}-\alpha_{1}=\frac{1}{3}\left(a_{99}-a_{100}\right)=0$.

Thus, all our numbers $a_{1}, a_{2}, \ldots, a_{99}, a_{100}$ are equal to each other; therefore, if $a_{1}=1$, then $a_{2}=a_{3}=\ldots=a_{100}=1$.