There are given $1000$ integers $a_1,... , a_{1000}$. Their squares $a^2_1, . . . , a^2_{1000}$ are written in a circle. It so happened that the sum of any $41$ consecutive numbers on this circle is a multiple of $41^2$. Is it necessarily true that every integer $a_1,... , a_{1000}$ is a multiple of $41$?

(Boris Frenkin)
1. **Understanding the Problem:**
   We are given 1000 integers \(a_1, a_2, \ldots, a_{1000}\). Their squares \(a_1^2, a_2^2, \ldots, a_{1000}^2\) are arranged in a circle. The sum of any 41 consecutive numbers on this circle is a multiple of \(41^2\). We need to determine if it is necessarily true that every integer \(a_1, a_2, \ldots, a_{1000}\) is a multiple of 41.

2. **Analyzing the Condition:**
   The condition states that the sum of any 41 consecutive squares is a multiple of \(41^2\). Mathematically, for any \(k\),
   \[
   a_k^2 + a_{k+1}^2 + \cdots + a_{k+40}^2 \equiv 0 \pmod{41^2}
   \]

3. **Using Modular Arithmetic:**
   Consider the sequence of squares modulo \(41^2\). Since the sum of any 41 consecutive terms is 0 modulo \(41^2\), we can write:
   \[
   a_k^2 + a_{k+1}^2 + \cdots + a_{k+40}^2 \equiv 0 \pmod{41^2}
   \]
   for all \(k\).

4. **Implication for Individual Terms:**
   Since this holds for any \(k\), it implies that the sum of any 41 consecutive squares is divisible by \(41^2\). This suggests a periodicity or a strong divisibility condition on the individual terms.

5. **Considering the Divisibility:**
   Let's consider the implications of the sum being divisible by \(41^2\). If \(a_i\) were not a multiple of 41, then \(a_i^2\) would not be divisible by \(41^2\). However, since the sum of 41 such terms is divisible by \(41^2\), each term must contribute a multiple of \(41^2\) to the sum.

6. **Conclusion:**
   Therefore, for the sum of any 41 consecutive squares to be divisible by \(41^2\), each \(a_i^2\) must itself be divisible by \(41^2\). This implies that each \(a_i\) must be divisible by 41, because if \(a_i\) were not divisible by 41, \(a_i^2\) would not be divisible by \(41^2\).

\[
\boxed{\text{True}}
\]