Recall exponential notation for partitions: $a^b$ signifies $b$ occurrences of $a$ in the partition. (Exponential notation can be useful for seeing the generating functions.) In exponential notation, every partition satisfying your constraints are of the form $m^k (m+1)^l$. In your $n = 4$ example, the partitions are $4^1 5^0$, $2^2 3^0$, $1^2 2^1$, and $1^4 2^0$. Notice that the smaller number, $a_1$, must have exponent at least $1$, while successor can have exponent $0$.
For a fixed $m$, the contributions from partitions of the form $m^k (m+1)^l$ are given by the following generating function:
$$(x^m + x^{2m} + x^{3m} + \cdots)(1 + x^{m+1} + x^{2(m+1)} + \cdots)$$
This simplifies to:
$$\frac{x^m}{1-x^m} \frac{1}{1-x^{m+1}}$$
Such contributions come from any $m \geq 1$, and of course, the contributions are disjoint. Thus the full generating function is:
$$\sum_{m \geq 1} \frac{x^m}{1-x^m} \frac{1}{1-x^{m+1}}$$
This might already be too great a nudge, but the point is that you now obtain something you can manipulate. After some obvious $1 - x$ factorings and some telescoping, I get $\frac{x}{(1 - x)^2}$, a generating function for $n$, as desired. Let me know if you get similar results or not.
The number has to be $15317$. I give in to the temptation of posting my earlier comment as an answer. The number ends in $17$, so we have a starting point, the digit sum is $8$. We now need a digit sum of $9$, and the number appended on the left has to be divisible by $9$. Since $17$ and $9$ are coprime, $17$ x $9$ = $153$ is the smallest such number. Hence the answer to the original question is $15317$.
Best Answer
Fix $n$. The cases $n \le 3$ can be handled directly. We now assume $n > 3$.
Let $m = \lceil n \log_{10} 5 \rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.