4. $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of $d$, then the necessary and sufficient condition for the sum of any (different) two terms in the sequence to still be a term in this sequence is ( ).
(A) There exists an integer $m \geqslant-1$, such that $a_{1}=m d$
(B) $a_{n}=0(n=1,2, \cdots)$
(C) $a_{1}$ is an integer, $d=1$
(D) $a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+a_{n}$
4. (A). We get $a_{4}+a_{t}=u_{k}$, so
$$
\begin{array}{l}
2 a_{1}+(s+t-2) d=a_{1}+(k-1) d . \\
\therefore a_{1}=(k-s-t+1) d .
\end{array}
$$

Therefore, there exists an integer $m=k-s-t+1$, such that $a_{1}=m d$.
Next, we prove that $m \geqslant-1$.
When $d \neq 0$, it is obviously true. For $d \neq 0$, if $m$ is less than -1, then $p=-m \geqslant 2$. For two different terms $a_{1}$ and $a_{n}$, there exists $a_{k}$ such that
$$
a_{1}+a_{p}=a_{k},
$$

which means $2 m d+(-m-1) d=m d+(k-1) d$.
Thus, $k d=0$. This contradicts $k>0, d \neq 0$.
Conversely, if there exists $m \geqslant-1$ such that $a_{1}=m d$, for $s \neq t$, we have
$$
\begin{array}{l}
s+t \geqslant 3, m+s+t \geqslant 2, \text { and } \\
a_{s}+a_{t}=2 a_{1}+(s+t-2) d \\
=a_{1}+(m+s+t-2) d=a_{m+s+t-1} .
\end{array}
$$

That is, there exists the $(m+s+t-1) \geqslant 1$-th term, which is exactly equal to the sum of $a_{s}+a_{t}$.
Option (B) is sufficient but not necessary; Option (C) is sufficient but not necessary when $a \geqslant-1$, and neither sufficient nor necessary when $a \leqslant-2$; Option (D) is not an arithmetic sequence.