1. The transformation $ n \to 2n \minus{} 1$ or $ n \to 3n \minus{} 1$, where $ n$ is a positive integer, is called the 'change' of $ n$. Numbers $ a$ and $ b$ are called 'similar', if there exists such positive integer, that can be got by finite number of 'changes' from both $ a$ and $ b$. Find all positive integers 'similar' to $ 2005$ and less than $ 2005$.
1. **Understanding the Problem:**
   We need to find all positive integers less than 2005 that are 'similar' to 2005. Two numbers \(a\) and \(b\) are 'similar' if there exists a positive integer that can be obtained from both \(a\) and \(b\) by a finite number of transformations \(n \to 2n - 1\) or \(n \to 3n - 1\).

2. **Claim 1:**
   Numbers divisible by 6 and those which are even and \(\equiv 2 \pmod{3}\) cannot be similar to a number smaller than it.
   - **Proof:**
     If a number \(n\) is divisible by 6, then \(n = 6k\) for some integer \(k\). Applying the transformations:
     \[
     6k \to 2(6k) - 1 = 12k - 1 \quad \text{or} \quad 6k \to 3(6k) - 1 = 18k - 1
     \]
     Both results are not divisible by 6, and thus cannot be similar to a smaller number that is divisible by 6.
     Similarly, if \(n\) is even and \(\equiv 2 \pmod{3}\), then \(n = 6k + 2\). Applying the transformations:
     \[
     6k + 2 \to 2(6k + 2) - 1 = 12k + 3 \quad \text{or} \quad 6k + 2 \to 3(6k + 2) - 1 = 18k + 5
     \]
     Both results are not even and \(\equiv 2 \pmod{3}\), and thus cannot be similar to a smaller number with the same properties.

3. **Claim 2 (Transitivity):**
   If \(a \sim b\) and \(b \sim c\), then \(a \sim c\).
   - **Proof:**
     If \(a\) can be transformed to \(b\) and \(b\) can be transformed to \(c\), then by combining the sequences of transformations, \(a\) can be transformed to \(c\). Hence, \(a \sim c\).

4. **Claim 3:**
   The transformations are all increasing, i.e., the transformations are not reflexive.
   - **Proof:**
     For \(n \to 2n - 1\):
     \[
     2n - 1 \geq n \quad \text{for all } n \geq 1
     \]
     For \(n \to 3n - 1\):
     \[
     3n - 1 \geq n \quad \text{for all } n \geq 1
     \]
     Therefore, the transformations always increase the value of \(n\).

5. **Finding Similar Numbers to 2005:**
   We need to find numbers less than 2005 that can be transformed to 2005 by reversing the transformations. Starting from 2005:
   \[
   2005 \to \frac{2005 + 1}{2} = 1003 \quad \text{(since } 2005 \equiv 1 \pmod{2})
   \]
   \[
   1003 \to \frac{1003 + 1}{2} = 502 \quad \text{(since } 1003 \equiv 1 \pmod{2})
   \]
   \[
   502 \to \frac{502 + 1}{3} = 167 \quad \text{(since } 502 \equiv 2 \pmod{3})
   \]
   \[
   167 \to \frac{167 + 1}{2} = 84 \quad \text{(since } 167 \equiv 1 \pmod{2})
   \]
   \[
   84 \to \frac{84 + 1}{3} = 28 \quad \text{(since } 84 \equiv 0 \pmod{3})
   \]
   \[
   28 \to \frac{28 + 1}{2} = 14 \quad \text{(since } 28 \equiv 0 \pmod{2})
   \]
   \[
   14 \to \frac{14 + 1}{2} = 7 \quad \text{(since } 14 \equiv 0 \pmod{2})
   \]
   \[
   7 \to \frac{7 + 1}{2} = 4 \quad \text{(since } 7 \equiv 1 \pmod{2})
   \]
   \[
   4 \to \frac{4 + 1}{3} = 1 \quad \text{(since } 4 \equiv 1 \pmod{3})
   \]

   Therefore, the numbers similar to 2005 and less than 2005 are \(1003, 502, 167, 84, 28, 14, 7, 4, 1\).

The final answer is \(\boxed{1003, 502, 167, 84, 28, 14, 7, 4, 1}\)