5.2899999 .

Let the smallest number in $A$ be $n=\overline{a_{1} a_{2} \cdots a_{t}}$,
$$
S(n)=a_{1}+a_{2}+\cdots+a_{t} \text {. }
$$

If the unit digit of $n$, $a_{t} \neq 9$, then
$$
\begin{array}{l}
n+1=\overline{a_{1} a_{2} \cdots a_{t-1} a_{t}^{\prime}}\left(a_{t}^{\prime}=a_{t}+1\right) . \\
\text { Hence } S(n+1)=S(n)+1 .
\end{array}
$$

By the problem statement,
$$
S(n)=11 p_{1}, S(n+1)=11 p_{2} \text {. }
$$

Thus, $11\left(p_{2}-p_{1}\right)=1$, which is a contradiction.
Therefore, $a_{t}=9$.
Assume the last $k$ digits of $n$ are all 9. Then
$$
\begin{array}{l}
S(n)=a_{1}+a_{2}+\cdots+a_{t-k}+9 k=11 p_{1}, \\
S(n+1)=a_{1}+a_{2}+\cdots+\left(a_{t-k}+1\right)+0 \times k \\
=11 p_{2} .
\end{array}
$$

From the above two equations, we get $11\left(p_{1}-p_{2}\right)=9 k-1$.
For the smallest $n$, there is the smallest natural number $k$ such that $111(9 k-1)$.
Upon inspection, $k=1,2,3,4$ do not meet the requirements.
When $k=5$, $9 \times 5-1=44$ is a multiple of 11, at this time, $p_{1}-p_{2}=4$.
Therefore, the last five digits of $n$ are all 9.
Take the smallest $n$ such that $p_{1}-p_{2}=4$, then
$$
\begin{array}{l}
p_{1}=5, p_{2}=1 \\
\Rightarrow S(n)=55, S(n+1)=11 .
\end{array}
$$

From $a_{1}+a_{2}+\cdots+a_{t-k}+9 \times 5=55$, we get
$$
a_{1}+a_{2}+\cdots+a_{t-k}=10>9 \Rightarrow t-k \geqslant 2 \text {. }
$$

Since $k=5$, the smallest $n$ is a seven-digit number, with the sum of the first two digits being 10.
Thus, $n=2899999, n+1=2900000$.
5. Given the set
$$
A=\{n|n \in \mathbf{N}, 11| S(n), 11 \mid S(n+1)\} \text {, }
$$

where $S(m)$ denotes the sum of the digits of the natural number $m$. Then the smallest number in set $A$ is $\qquad$ .