4. In an arithmetic sequence with real number terms, the common difference is 4, and the square of the first term plus the sum of the remaining terms does not exceed 100. Such a sequence can have at most  terms.
4.8 .

Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{aligned}
& a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2} \cdot(n-1) \\
& \leqslant 100 \\
\Leftrightarrow & a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{aligned}
$$

There exists at least one real number $a_{1}$ satisfying inequality (1) if and only if $\Delta \dot{=}^{\prime}(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0$. Since $\Delta \geqslant 0 \Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0$
$$
\Leftrightarrow n_{1} \leqslant n \leqslant n_{2} \text {, }
$$

where $n_{1}=\frac{3-\sqrt{2816}}{7}<0,8<n_{2}=\frac{3+\sqrt{2816}}{7}<9$.
Therefore, the maximum natural number $n$ satisfying inequality (2) is 8, meaning the sequence can have at most 8 terms.