2. Let $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$ be $n$ real numbers satisfying the following condition: for any integer $k>0$, we have $a_{1}^{k}+a_{2}^{k}+$ $\cdots+a_{n}^{k} \geqslant 0$. Then, $p=\max \left\{\left|a_{1}\right|,\left|a_{2}\right|,\right.$ $\left.\cdots,\left|a_{n}\right|\right\}=$ $\qquad$
2. $a_{1}$.

Obviously, $p$ is either equal to $a_{1}$, or equal to $\left|a_{n}\right|$.
Assume $p \neq a_{1}$, then $p=\left|a_{n}\right|$, and $p>a_{1}, a_{n}a_{n-k+1}=$ $a_{n-k+2}=\cdots=a_{n}$, then $\left|\frac{a_{1}}{a_{n}}\right|0 .
\end{array}
$$

Thus, $a_{1}^{2 l+1}+\cdots+a_{n}^{2 l+1}$
$$
\begin{array}{l}
=a_{n}^{2 l+1} \cdot\left[\left(\frac{a_{1}}{a_{n}}\right)^{2 l+1}+\cdots+\left(\frac{a_{n}}{a_{n}}\right)^{2 l+1}\right] \\
<0 .
\end{array}
$$

This contradicts the given information.
Therefore, it must be that $p=a_{1}$.