There is infinite sequence of composite numbers $a_1,a_2,...,$ where $a_{n+1}=a_n-p_n+\frac{a_n}{p_n}$ ; $p_n$ is smallest prime divisor of $a_n$. It is known, that $37|a_n$ for every $n$.
Find possible values of $a_1$
1. Given the sequence \(a_1, a_2, \ldots\) where \(a_{n+1} = a_n - p_n + \frac{a_n}{p_n}\) and \(p_n\) is the smallest prime divisor of \(a_n\), and it is known that \(37 \mid a_n\) for every \(n\).

2. We need to find the possible values of \(a_1\).

3. Assume there exists an \(n\) such that \(p_n \neq 37\). Since \(37 \mid a_n\), we have:
   \[
   37 \mid a_n \quad \text{and} \quad 37 \mid \frac{a_n}{p_n}
   \]
   This implies:
   \[
   37 \mid a_{n+1} = a_n - p_n + \frac{a_n}{p_n}
   \]
   Therefore, \(37 \mid p_n\), which implies \(p_n = 37\). This is a contradiction since \(p_n\) is the smallest prime divisor of \(a_n\) and cannot be 37 if \(p_n \neq 37\).

4. Hence, \(p_n = 37\) for all \(n \in \mathbb{Z}^+\).

5. Let \(a_n = 37b_n\). Then:
   \[
   a_{n+1} = 37b_n - 37 + \frac{37b_n}{37} = 37b_n - 37 + b_n = 37(b_n - 1 + \frac{b_n}{37})
   \]
   This implies:
   \[
   b_{n+1} = b_n - 1 + \frac{b_n}{37}
   \]
   Since \(37 \mid b_n\), let \(b_n = 37c_n\). Then:
   \[
   b_{n+1} = 37c_n - 1 + \frac{37c_n}{37} = 37c_n - 1 + c_n = 37c_n + c_n - 1 = 38c_n - 1
   \]
   Therefore:
   \[
   c_{n+1} = c_n - \frac{1}{37}
   \]
   Since \(37 \mid c_n - 1\), let \(c_n = 37d_n + 1\). Then:
   \[
   c_{n+1} = 37d_n + 1 - \frac{1}{37} = 37d_n + \frac{36}{37}
   \]
   This implies:
   \[
   37d_{n+1} = 38d_n
   \]
   Therefore:
   \[
   d_{n+1} = \frac{38}{37}d_n
   \]
   Since \(d_n\) must be an integer, and \(37^{\ell} \mid d_n\) for all \(n \in \mathbb{Z}^+\) and all \(\ell \in \mathbb{Z}^+\), we conclude that \(d_n = 0\) for all \(n \in \mathbb{Z}^+\).

6. Hence, \(a_n = 37^2(37d_n + 1) = 37^2\).

The final answer is \(\boxed{37^2}\).