5. Let $x_{1}, x_{2}, \cdots, x_{51}$ be natural numbers, $x_{1}<x_{2}$ $<\cdots<x_{51}$, and $x_{1}+x_{2}+\cdots+x_{51}=1995$. When $x_{26}$ reaches its maximum value, the maximum value that $x_{51}$ can take is
5. To maximize $x_{26}$, then $x_{1}, x_{2}, \cdots, x_{25}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+x_{27}+\cdots+x_{51} \\
=1995-(1+2+\cdots+25)=1670 .
\end{array}
$$

To maximize $x_{51}$, then $x_{26}, x_{27}, \cdots, x_{50}$ must be minimized, thus we have
$$
\begin{array}{l}
x_{26}+\left(x_{26}+1\right)-1-\cdots+\left(x_{26}+24\right)+x_{01} \\
=1670\left(x_{11}>x_{23}+\% 4\right),
\end{array}
$$

i.e. $\square$
$$
\begin{array}{l}
25 x_{26}+-x_{51}=1670-(1+2+\cdots+24) \\
=1370=25 \times 51+95 .
\end{array}
$$

Therefore, the maximum value that $x_{51}$ can achieve is 95.