4. Let $a_{1}=1, S_{n+1}=2 S_{n}-\frac{n(n+1)}{2}+1$, where $S_{n}=a_{1}+a_{2}+\cdots+a_{n}(n=1,2, \cdots)$. If we define $\Delta a_{n}=a_{n+1}-a_{n}$, then the set
$$
S=\left\{n \mid n \in \mathbf{N}_{+}, \Delta\left(\Delta a_{n}\right) \geqslant-2006\right\}
$$

has the number of elements $|S|$ as ( ).
(A)9
(B) 10
(C) 11
(D) 12
4.C.

From the problem, we have
$$
\begin{array}{l}
S_{n+1}=2 S_{n}-\frac{n(n+1)}{2}+1, \\
S_{n}=2 S_{n-1}-\frac{(n-1) n}{2}+1(n \geqslant 2) .
\end{array}
$$
Subtracting (2) from (1) gives
$$
a_{n+1}=2 a_{n}-n(n \geqslant 2) \text {. }
$$

Taking $n=1$ in the given equation, we get
$$
a_{1}+a_{2}=2 a_{1}-1+1 \text {, }
$$

which simplifies to $a_{2}=a_{1}=1$.
From equation (3), we have
$$
a_{n+1}=2 a_{n}-n(n \geqslant 1) .
$$

Thus, $a_{n+1}-(n+2)=2\left(a_{n}-n-1\right)$.
Let $a_{n}-n-1=b_{n}$, then
$$
b_{n+1}=2 b_{n}, b_{1}=a_{1}-2=-1 \text {. }
$$

Therefore, $b_{n}=-2^{n-1}(n \geqslant 1)$.
So, $a_{n}=-2^{n-1}+n+1(n \geqslant 1)$.
By definition, $\Delta a_{n}=a_{n+1}-a_{n}$, we get
$$
\begin{array}{l}
\Delta\left(\Delta a_{n}\right)=\Delta a_{n+1}-\Delta a_{n} \\
=\left(a_{n+2}-a_{n+1}\right)-\left(a_{n+1}-a_{n}\right) \\
=\left(a_{n+1}-n-1\right)-\left(a_{n}-n\right) \\
=\left(-2^{n}+1\right)-\left(-2^{n-1}+1\right)=-2^{n-1} .
\end{array}
$$

Let $-2^{n-1} \geqslant-2006$, we get $1 \leqslant n \leqslant 11$, hence $|S|=11$.