There are $N{}$ mess-loving clerks in the office. Each of them has some rubbish on the desk. The mess-loving clerks leave the office for lunch one at a time (after return of the preceding one). At that moment all those remaining put half of rubbish from their desks on the desk of the one who left. Can it so happen that after all of them have had lunch the amount of rubbish at the desk of each one will be the same as before lunch if a) $N = 2{}$ and b) $N = 10$?

[i]Alexey Zaslavsky[/i]
1. **Initial Setup:**
   Let the $N$ mess-loving clerks have $a_1, a_2, \cdots, a_N$ amounts of rubbish on their desks initially.

2. **Case \(N = 2\):**
   - Let the initial amounts of rubbish be $a_1$ and $a_2$.
   - When the first clerk leaves, the second clerk puts half of their rubbish on the first clerk's desk. The new amounts of rubbish are:
     \[
     a_1' = a_1 + \frac{a_2}{2}, \quad a_2' = \frac{a_2}{2}
     \]
   - When the second clerk leaves, the first clerk puts half of their rubbish on the second clerk's desk. The new amounts of rubbish are:
     \[
     a_1'' = \frac{a_1'}{2}, \quad a_2'' = a_2' + \frac{a_1'}{2}
     \]
   - Substituting $a_1'$ and $a_2'$:
     \[
     a_1'' = \frac{a_1 + \frac{a_2}{2}}{2} = \frac{a_1}{2} + \frac{a_2}{4}
     \]
     \[
     a_2'' = \frac{a_2}{2} + \frac{a_1 + \frac{a_2}{2}}{2} = \frac{a_2}{2} + \frac{a_1}{2} + \frac{a_2}{4} = \frac{a_1}{2} + \frac{3a_2}{4}
     \]
   - Equating the final amounts of rubbish with the original:
     \[
     a_1 = \frac{a_1}{2} + \frac{a_2}{4}
     \]
     \[
     a_2 = \frac{a_1}{2} + \frac{3a_2}{4}
     \]
   - Solving these equations:
     \[
     4a_1 = 2a_1 + a_2 \implies 2a_1 = a_2
     \]
     \[
     4a_2 = 2a_1 + 3a_2 \implies a_2 = 2a_1
     \]
   - Both equations are consistent, confirming that $2a_1 = a_2$.

3. **Case \(N = 10\):**
   - Consider the construction $(1, 2, 4, 8, \cdots, 2^9)$.
   - When the first clerk leaves, each of the remaining clerks puts half of their rubbish on the first clerk's desk. The new amounts of rubbish are:
     \[
     a_1' = 1 + \frac{2}{2} + \frac{4}{2} + \cdots + \frac{2^9}{2} = 1 + 1 + 2 + 4 + \cdots + 2^8
     \]
     \[
     a_2' = \frac{2}{2}, \quad a_3' = \frac{4}{2}, \quad \cdots, \quad a_{10}' = \frac{2^9}{2}
     \]
   - This process cycles the numbers, and after $N$ moves, we get the same sequence $(1, 2, 4, 8, \cdots, 2^9)$.
   - Therefore, the amounts of rubbish on each desk remain the same after all clerks have had lunch.

The final answer is \( \boxed{ 2a_1 = a_2 } \) for \(N = 2\) and \((1, 2, 4, 8, \cdots, 2^9)\) for \(N = 10\).