Find the four smallest four-digit numbers that meet the following condition: by dividing by $2$, $3$, $4$, $5$ or $6$ the remainder is $ 1$.
1. **Set up the congruences:**
   We need to find the smallest four-digit numbers \( x \) such that:
   \[
   \begin{align*}
   x &\equiv 1 \pmod{2}, \\
   x &\equiv 1 \pmod{3}, \\
   x &\equiv 1 \pmod{4}, \\
   x &\equiv 1 \pmod{5}, \\
   x &\equiv 1 \pmod{6}.
   \end{align*}
   \]

2. **Combine the congruences:**
   Notice that all the moduli (2, 3, 4, 5, 6) are factors of 60. Therefore, we can combine these congruences into a single congruence:
   \[
   x \equiv 1 \pmod{60}.
   \]
   This is because the least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60.

3. **Express \( x \) in terms of \( n \):**
   From the combined congruence, we have:
   \[
   x = 60n + 1,
   \]
   where \( n \) is a natural number.

4. **Find the smallest four-digit numbers:**
   We need to find the smallest values of \( n \) such that \( x \) is a four-digit number. A four-digit number ranges from 1000 to 9999.

   \[
   1000 \leq 60n + 1 \leq 9999
   \]

   Solving for \( n \):
   \[
   1000 \leq 60n + 1 \implies 999 \leq 60n \implies n \geq \frac{999}{60} \approx 16.65
   \]
   Since \( n \) must be an integer, we take the next whole number:
   \[
   n \geq 17
   \]

5. **Calculate the smallest four-digit numbers:**
   For \( n = 17 \):
   \[
   x = 60 \cdot 17 + 1 = 1021
   \]

   For \( n = 18 \):
   \[
   x = 60 \cdot 18 + 1 = 1081
   \]

   For \( n = 19 \):
   \[
   x = 60 \cdot 19 + 1 = 1141
   \]

   For \( n = 20 \):
   \[
   x = 60 \cdot 20 + 1 = 1201
   \]

   Therefore, the four smallest four-digit numbers that meet the given condition are:
   \[
   \boxed{1021, 1081, 1141, 1201}
   \]