Solution:

Consider a 6-digit number whose digits from left to right are in non-increasing order. If $1$ is the first digit of such a number, then the subsequent digits cannot exceed $1$. The set of all such numbers with initial digit equal to $1$ is
$$\{100000, 110000, 111000, 111100, 111110, 111111\}.$$
There are $6$ elements in this set.

Let us consider 6-digit numbers with initial digit $2$. Starting from $200000$, we can go up to $222222$. We count these numbers as follows:

The number of such numbers is $21$.

Similarly we count numbers with initial digit $3$; the sequence starts from $300000$ and ends with $333333$. We have

We obtain the total number of numbers starting from $3$ equal to $56$.

Similarly,

Thus the number of 6-digit numbers where digits are non-increasing starting from $100000$ and ending with $777777$ is
$$792 + 462 + 252 + 126 + 56 + 21 + 6 = 1715$$
Since $2005 - 1715 = 290$, we have to consider only $290$ numbers in the sequence with initial digit $8$. We have
$$\begin{aligned} & 800000 - 855555 \quad: \quad 252 \\ & 860000 - 863333: 35 \\ & 864000 - 864110 \quad: \quad 3 \end{aligned}$$
Thus the required number is $\underline{864110}$.

It is known that the number of ways of choosing $r$ objects from $n$ different types of objects (with repetitions allowed) is $\binom{n+r-1}{r}$. In particular, if we want to write $r$-digit numbers using $n$ digits allowing for repetitions with the additional condition that the digits appear in non-increasing order, we see that this can be done in $\binom{n+r-1}{r}$ ways.

Now we group the given numbers into different classes and write the number of ways in which each class can be obtained. To keep track we also write the cumulative sums of the number of numbers so obtained. Observe that the numbers themselves are written in ascending order. So we exhaust numbers beginning with $1$, then beginning with $2$ and so on.

The next three 6-digit numbers are $864000, 864100, 864110$.

Hence the $2005$-th number in the sequence is $864110$.