The general solution is
$$n = \frac{10}{91} \times (10^{6k} - 1) = (1 + 10^6 + 10^{12} + \dots + 10^{6(k-1)}) \times 109890$$
for any positive integer $k$. In other words, the number $n$ consists of $k$ replicates of the digits 109890.

Let $m$ be the number of digits of $n$ so $m \ge 3$ when $n \ge 100$. Dividing $n$ by $10^{m-2}$, let $q$ be the quotient and $r$ the remainder. This splits the digits of $n$ into the first two and the remaining digits,
$$n = 10^{m-2}q + r$$
where $10 \le q \le 99$ and $0 \le r < 10^{m-2}$. Rearranging the digits we have:
$$T(n) = 100r + q.$$
The equation $n + T(n) = 10n$ is equivalent to $T(n) = 9n$ which we write in terms of $q$ and $r$:
$$100r + q = 9 \times 10^{m-2}q + 9r$$
or
$$91r = (9 \times 10^{m-2} - 1)q. \qquad (7)$$

We know that $q \ge 10$. We now claim that $q = 10$ for any solution. For a contradiction, suppose that $q \ge 11$. Then equation (7) implies that
$$(9 \times 10^{m-2} - 1) \times 11 \le (9 \times 10^{m-2} - 1)q = 91r < 91 \times 10^{m-2}.$$
Comparing the left and right end gives $(99-91) \times 10^{m-2} \le 11$, contradicting $m \ge 3$. Therefore $q = 10$. Substituting $q = 10$ into equation (7) gives
$$91r = (9 \times 10^{m-2} - 1) \times 10.$$
As $\gcd(91, 10) = 1$ then 91 must divide $9 \times 10^{m-2} - 1$. This is not only necessary but also sufficient, as we can then write:
$$r = \frac{9 \times 10^{m-2} - 1}{91} \times 10 < \frac{9}{91} \times 10^{m-1} < 10^{m-2} \quad (8)$$
and $r$ is then an integer that lies in the feasible range.
It remains then to find the values of $m$ for which $9 \times 10^{m-2} - 1$ is a multiple of 91. Equivalently, we require that $(9 + 91) \times 10^{m-2} - 1 = 10^m - 1$, is a multiple of 91. As $91 = 7 \times 13$ we can use a short cut to check that $10^m - 1$ is a multiple both of 7 and of 13. Remainders tabulated below:

The equation $n + T(n) = 10n$ could also be written as $T(n) = 9n$.
As $n$ and $T(n)$ have the same number of digits, the leading digit of $n$ must be equal to 1 and the leading digit of $T(n) = 9n$, which is the third digit from the left of $n$, has then to be equal to 9. This implies that there occurs a carry from the third to the second place from the left, hence the second digit of $n$ must be zero. The shapes of $n$ and $T(n)$ are therefore
$$n = 109 **\ldots** * \qquad T(n) = 9 **\ldots** * 10$$
where each star stands for a digit and the dots indicate that the number of digits of $n$ is currently unknown. We can now work our way backwards from $T(n) = 9n$ to $n$, starting at the end. The key feature we are using here is that $9k \equiv -k \equiv 10 - k \pmod{10}$.
Let the rightmost unknown digit be equal to $d$ so that we have
$$n = 109 **\ldots** * d \qquad 9n = T(n) = 9 **\ldots** * d10$$
then $9d \equiv 0 \pmod{10}$, hence $d = 0$. This gives us
$$n = 109 **\ldots** * 0 \qquad 9n = T(n) = 9 **\ldots** * 010$$
We will again call the rightmost unknown digit $d$ so that we have
$$n = 109 **\ldots** * d0 \qquad 9n = T(n) = 9 **\ldots** * d010$$
and $9d \equiv 1 \pmod{10}$, hence $d = 9$. We now need to check if this digit could already be the third digit from the left, i.e. $n = 1090$. But $1090 \cdot 9 = 9810 \neq T(1090)$. This gives us, with new $d$,
$$n = 109 **\ldots** * d90 \qquad 9n = T(n) = 9 **\ldots** * d9010$$
and, keeping the carry in mind, $9d + 8 \equiv 0 \pmod{10}$. Therefore, $d = 8$ and
$$n = 109 **\ldots** * d890 \qquad 9n = T(n) = 9 **\ldots** * d89010$$
with $9d + 8 \equiv 9 \pmod{10}$, which implies $d = 9$. We have to check again if this digit could already be the third digit from the left, i.e. $n = 109890$. Indeed, we now have $9n = 9 \cdot 109890 = 989010 = T(n)$.
So far we have shown that $109890$ is one solution and that the last six digits of $T(n)$ for any solution $n$ need to be $989010$. This implies that the last six digits of any solution $n$ are equal to $989010/9 = 109890$.
We could now continue in the manner above to see that this sequence of digits must repeat if $n$ has more than six digits. Alternatively, we could argue as follows.

Let $m$ be the number obtained from $n$ by removing the last six digits. We show that $T(m) = 9m$. Clearly, $T(m)$ agrees, except possible at the last two digits, with the number $T'(n)$ obtained from $T(n)$ by removing the last six digits. As we have seen above, the two leading digits of $n$, and so also of $m$, are 10. As the last six digits of $n$ are 109890, the last two digits of $T'(n)$ are 10. Hence, $T'(n) = T(m)$. Using $T(n) = 9n$ it follows now that $T(m) = 9m$. It follows by induction that $n$ has $6k$ digits ($k \ge 1$) and consists of $k$ consecutive blocks of six digits, all equal to 109890.