Find all real constants c for which there exist strictly increasing sequence $a$ of positive integers such that $(a_{2n-1}+a_{2n})/{a_n}=c$ for all positive intеgers n.
1. **Establishing the lower bound for \( c \):**

   Consider the sequence \( \Delta_n = a_{n+1} - a_n \) for convenience, and denote \( \Delta_0 = 0 \). Given the condition \( \frac{a_{2n-1} + a_{2n}}{a_n} = c \), we can rewrite it as:
   \[
   a_{2n-1} + a_{2n} = c a_n
   \]
   Similarly, for \( n+1 \):
   \[
   a_{2n+1} + a_{2n+2} = c a_{n+1}
   \]
   Subtracting these two equations, we get:
   \[
   (a_{2n+1} + a_{2n+2}) - (a_{2n-1} + a_{2n}) = c (a_{n+1} - a_n)
   \]
   Using the definition of \( \Delta \), we can express this as:
   \[
   \Delta_{2n-1} + 2\Delta_{2n} + \Delta_{2n+1} = c \Delta_n
   \]
   To find the minimum value of \( c \), note that the minimum of \( \Delta_{2n-1}, \Delta_{2n}, \Delta_{2n+1} \) must not exceed \( \frac{c}{4} \Delta_n \). Hence, there exists an \( i_1 \) such that:
   \[
   \Delta_{i_1} \leq \frac{c}{4} \Delta_n
   \]
   Continuing this process, we find a sequence \( (i_k) \) such that:
   \[
   \Delta_{i_k} \leq \left( \frac{c}{4} \right)^k \Delta_n
   \]
   If \( c < 4 \), then \( \left( \frac{c}{4} \right)^k \) converges to 0 as \( k \to \infty \), which is impossible since \( \Delta_n \) are positive integers. Therefore, \( c \geq 4 \).

2. **Constructing a sequence for \( c = 4 \):**

   Consider the sequence of odd natural numbers: \( 1, 3, 5, 7, \ldots \). For this sequence:
   \[
   a_{2n-1} + a_{2n} = (2n-1) + (2n+1) = 4n = 4 \cdot n
   \]
   Thus, \( c = 4 \) works.

3. **Proving \( c \) must be an integer:**

   Assume \( c \) is not an integer, i.e., \( c = \frac{p}{q} \) with \( q > 2 \) and \( p, q \) are coprime. Using the equation:
   \[
   \frac{p}{q} \Delta_n = \Delta_{2n-1} + 2\Delta_{2n} + \Delta_{2n+1}
   \]
   Since the right-hand side is an integer, \( q \) must divide \( \Delta_n \) for all \( n \). Continuing this, we must have \( q^k \mid \Delta_n \) for all \( k \), which is impossible for \( q \geq 2 \). Hence, \( c \) must be an integer.

4. **Constructing a sequence for \( c > 4 \):**

   For \( c \geq 5 \), we can construct a sequence as follows:
   \[
   a_1 = 1, \quad a_2 = c, \quad a_3 = \frac{c(c-1)}{2} - 1, \quad a_4 = \frac{c(c-1)}{2} + 1
   \]
   This sequence satisfies \( a_1 < a_2 < a_3 < a_4 \). If \( a_n \) has been constructed, there are two cases:
   - If \( ca_n \) is odd, take \( a_{2n-1} = \frac{ca_n - 1}{2} \) and \( a_{2n} = \frac{ca_n + 1}{2} \).
   - If \( ca_n \) is even, take \( a_{2n-1} = \frac{ca_n}{2} - 1 \) and \( a_{2n} = \frac{ca_n}{2} + 1 \).

   We need to check that \( (a_n) \) is strictly increasing. By induction, assume \( a_{2n-1} \leq a_{2n-2} \). Since \( a_{2n-1} \geq \frac{ca_n}{2} - 1 \) and \( a_{2n-2} \leq \frac{ca_{n-1}}{2} + 1 \), we have:
   \[
   c(a_n - a_{n-1}) \geq 4
   \]
   Since \( c \geq 5 \) and \( a_n - a_{n-1} \geq 1 \), this is a contradiction. Thus, \( a_{2n-1} > a_{2n-2} \).

   Therefore, the answer is all positive integers \( c \geq 4 \).

The final answer is \( \boxed{ c \geq 4 } \) and \( c \) is a natural number.