1, 3, 9, 27, 37, 67
Let us use the following notations in the subsequent argument to obtain the solution to this problem:
* For a pair of positive integers $(a, b)$, denote by $\gcd(a, b)$ the greatest common divisor of $a$ and $b$.
* For integers $a, b$ and $p$, we write $a \equiv b \pmod p$ to mean that $a - b$ is a multiple of $p$.
We also quote, without proof, the following well-known facts:
* When $a^n \equiv 1 \pmod p$ holds for positive integers $a, n, p$, the following statement holds for a positive integer $m$:
$$a^m \equiv 1 \pmod p \iff a^{\gcd(m,n)} \equiv 1 \pmod p.$$
* If $p$ is a prime, and an integer $a$ is not a multiple of $p$, then
$$a^{p-1} \equiv 1 \pmod p$$
holds (Fermat's Little Theorem).
Starting the search for the positive factors of $10^{2013} - 1$ less than or equal to 100, let us first determine the prime numbers $p$ less than 100, which are factors of $10^{2013} - 1$. From what we stated above it follows that we have
$$10^{2013} - 1 \equiv 0 \pmod p \iff 10^{2013} \equiv 1 \pmod p \iff 10^d \equiv 1 \pmod p,$$
where $d = \gcd(2013, p-1)$. Then, we see that $d$ is a factor of $2013 = 3 \cdot 11 \cdot 61$ and must satisfy $d \le p-1 < 99$. Therefore, we conclude that $d \in \{1, 3, 11, 33, 61\}$ must be satisfied. Let us go through the search for $p$ case by case depending on the possible value of $d$.
(1) When $d=1$: in this case, we see that $p=3$ is the only prime which satisfies $10^1 \equiv 1 \pmod p$, and it satisfies $\gcd(2013, p-1) = 1$ as well.
(2) When $d=3$: in this case, we have $10^3 - 1 = 3^3 \cdot 37$. So, $p=3, 37$ are the only primes satisfying $10^3 \equiv 1 \pmod p$. Of these two primes, 37 is the only one satisfying $\gcd(2013, p-1) = 3$.
(3) When $d=11$: in this case, $p=23, 89$ are the only primes less than 100 satisfying the condition $\gcd(2013, p-1) = 11$. For each of these two primes $p$, let us check whether it satisfies the condition $10^{11} \equiv 1 \pmod p$. Let $p=23$, then since we have
$$10^5 \equiv 10^2 \cdot 10^2 \cdot 10 \equiv 8 \cdot 8 \cdot 10 \equiv -4 \pmod{23},$$
from which it follows that
$$10^{11} \equiv 10^5 \cdot 10^5 \cdot 10 \equiv (-4) \cdot (-4) \cdot 10 \equiv -1 \pmod{23}.$$
Therefore, $10^{11} \equiv 1$ is not satisfied.
If $p=89$, then from
$$10^5 \equiv 10^2 \cdot 10^2 \cdot 10 \equiv 11 \cdot 11 \cdot 10 \equiv -36 \pmod{89}$$
it follows that
$$10^{11} \equiv 10^5 \cdot 10^5 \cdot 10 \equiv (-36) \cdot (-36) \cdot 10 \equiv 55 \pmod{89}.$$
Therefore, $10^{11} \equiv 1 \pmod{89}$ is not satisfied.
Consequently, there are no prime factors less than 100 for $10^{2013} - 1$ for the case $d=11$.
(4) When $d=33$: in this case, we see that 67 is the only prime less than 100 satisfying the condition $\gcd(2013, p-1) = 33$. Let us check whether $10^{33} \equiv 1 \pmod{67}$. By repeating to take squares, we get
$$10^2 \equiv 33, \quad 10^4 \equiv 17, \quad 10^8 \equiv 21, \quad 10^{16} \equiv 39, \quad 10^{32} \equiv 47 \pmod{67},$$
and by multiplying further by 10, we obtain $10^{33} \equiv 1 \pmod{67}$. Thus, we get $p=67$ is the prime factor of $10^{2103}-1$ corresponding to the case $d=33$.
(5) When $d=61$: in this case, we see that there is no prime $p$ less than 100 satisfying the condition $\text{gcd}(2013, p-1) = 61$.
Thus we conclude that only prime factors of $10^{2013}-1$ less than 100 are 3, 37, 67, and since any positive factor of $10^{2013}-1$ less than or equal to 100 cannot have any other prime factors, we see that only possible positive factors of $10^{2013}-1$ must come from
1, 3, 9, 27, 37, 67, 81.
We already checked that 3, 37, 67 all divide $10^{2013}-1$ and 1 is obviously a factor. So, it is enough to check whether any of 9, 27, 81 divides $10^{2013}-1$. Since we have
$$10^{2013}-1 = (10-1)(10^{2012} + 10^{2011} + \cdots + 10^1 + 1)$$
and
$$10^{2012} + 10^{2011} + \cdots + 10^1 + 1 \equiv 1^{2012} + 1^{2011} + \cdots + 1^1 + 1 \equiv 2013 \equiv 6 \pmod{9},$$
we get $10^{2013}-1 \equiv 54 \pmod{81}$. From this we conclude that both 9 and 27 are factors of $10^{2013}-1$, while 81 is not, and we get 1, 3, 9, 27, 37, 67 for the answer to the problem.