Determine the positive numbers $a{}$ for which the following statement true: for any function $f:[0,1]\to\mathbb{R}$ which is continuous at each point of this interval and for which $f(0)=f(1)=0$, the equation $f(x+a)-f(x)=0$ has at least one solution.

[i]Proposed by I. Yaglom[/i]
1. **Understanding the Problem:**
   We need to determine the positive numbers \( a \) for which the equation \( f(x+a) - f(x) = 0 \) has at least one solution for any continuous function \( f: [0,1] \to \mathbb{R} \) with \( f(0) = f(1) = 0 \).

2. **Reformulating the Problem:**
   The problem can be interpreted as finding \( a \) such that for any continuous function \( f \) satisfying the given conditions, there exists \( x \in [0, 1] \) such that \( f(x+a) = f(x) \).

3. **Using the Intermediate Value Theorem:**
   Since \( f \) is continuous on \([0,1]\) and \( f(0) = f(1) = 0 \), by the Intermediate Value Theorem, \( f \) must take every value between its minimum and maximum on \([0,1]\).

4. **Periodic Functions and the Universal Chord Theorem:**
   The Universal Chord Theorem states that for any continuous function \( f \) on \([0,1]\) with \( f(0) = f(1) \), there exists a \( c \in [0, 1] \) such that \( f(c) = f(c + \frac{1}{n}) \) for some positive integer \( n \).

5. **Applying the Universal Chord Theorem:**
   According to the Universal Chord Theorem, for any continuous function \( f \) on \([0,1]\) with \( f(0) = f(1) \), there exists a \( c \in [0, 1] \) such that \( f(c) = f(c + \frac{1}{n}) \) for some positive integer \( n \). This implies that \( a = \frac{1}{n} \) for some positive integer \( n \) will always satisfy the condition \( f(x+a) = f(x) \).

6. **Conclusion:**
   Therefore, the set of positive numbers \( a \) for which the equation \( f(x+a) - f(x) = 0 \) has at least one solution for any continuous function \( f: [0,1] \to \mathbb{R} \) with \( f(0) = f(1) = 0 \) is given by \( a = \frac{1}{n} \) for some positive integer \( n \).

The final answer is \( \boxed{ a = \frac{1}{n} } \) for some positive integer \( n \).