4. For 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 $\qquad$ terms.
8
4. 【Analysis and Solution】Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence with a common difference of 4, then
$$
\begin{array}{l}
a_{1}^{2}+a_{2}+a_{3}+\cdots+a_{n} \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+\frac{\left(a_{1}+4\right)+\left[a_{1}+4(n-1)\right]}{2}(n-1) \leqslant 100 \\
\Leftrightarrow a_{1}^{2}+(n-1) a_{1}+\left(2 n^{2}-2 n-100\right) \leqslant 0 .
\end{array}
$$

When and only when $\Delta=(n-1)^{2}-4\left(2 n^{2}-2 n-100\right) \geqslant 0$, there is at least no real number $a_{1}$ that satisfies inequality (1).
Since $\Delta \geqslant 0 \Leftrightarrow 7 n^{2}-6 n-401 \leqslant 0 \Leftrightarrow n_{1} \leqslant n \leqslant n^{2}$,
where $n_{1}=\frac{3-\sqrt{2816}}{7}<0,8<n^{2}=\frac{3+\sqrt{2816}}{7}<9$.
Therefore, the maximum value of the natural number $n$ that satisfies inequality (2) is 8, that is, the sequence that meets the problem's conditions has at most 8 terms.