Find all positive integers $N$ having only prime divisors $2,5$ such that $N+25$ is a perfect square.
1. We start with the equation given in the problem:
   \[
   N + 25 = M^2
   \]
   This implies:
   \[
   N = M^2 - 25 = (M+5)(M-5)
   \]
   Since \(N\) has only prime divisors 2 and 5, both \(M+5\) and \(M-5\) must be products of powers of 2 and 5.

2. Consider the case where \(5 \nmid M\). In this case, both \(M+5\) and \(M-5\) must be powers of 2. However, the difference between two powers of 2 is never 10 (since \(M+5 - (M-5) = 10\)). Therefore, \(5 \mid M\).

3. Let \(M = 5K\). Then:
   \[
   N = (5K+5)(5K-5) = 25(K+1)(K-1)
   \]
   Since \(N\) has only prime divisors 2 and 5, \((K+1)(K-1)\) must also be a product of powers of 2 and 5.

4. Consider the case where \(5 \nmid (K+1)(K-1)\). This implies that both \(K+1\) and \(K-1\) are powers of 2. Let \(K-1 = 2^a\) and \(K+1 = 2^b\) with \(b > a\). Then:
   \[
   2^b - 2^a = 2
   \]
   This simplifies to:
   \[
   2^a(2^{b-a} - 1) = 2
   \]
   Since \(2^a\) must be a power of 2, \(a = 1\) and \(2^{b-a} - 1 = 1\), so \(b-a = 1\). Therefore, \(a = 1\) and \(b = 2\), giving \(K-1 = 2\) and \(K+1 = 4\), so \(K = 3\). Thus:
   \[
   N = 25(K+1)(K-1) = 25 \cdot 4 \cdot 2 = 200
   \]

5. Now consider the case where \(5 \mid (K+1)(K-1)\). We have two subcases:
   - \(5 \mid K+1\): Let \(K+1 = 5m\). Then \(K-1 = 5m - 2\). Since \(K-1\) is a power of 2, let \(K-1 = 2^a\). Then:
     \[
     5m - 2 = 2^a
     \]
     and:
     \[
     K+1 = 5m
     \]
     Substituting \(K = 5m - 1\), we get:
     \[
     5m - 1 - 1 = 2^a \implies 5m - 2 = 2^a
     \]
     Since \(5m - 2\) is a power of 2, let \(5m - 2 = 2^a\). Then:
     \[
     5m = 2^a + 2
     \]
     For \(a = 3\), \(2^a + 2 = 10\), so \(m = 2\). Thus \(K = 9\), and:
     \[
     N = 25(K+1)(K-1) = 25 \cdot 10 \cdot 8 = 2000
     \]

   - \(5 \mid K-1\): Let \(K-1 = 5n\). Then \(K+1 = 5n + 2\). Since \(K+1\) is a power of 2, let \(K+1 = 2^b\). Then:
     \[
     5n + 2 = 2^b
     \]
     and:
     \[
     K-1 = 5n
     \]
     Substituting \(K = 5n + 1\), we get:
     \[
     5n + 1 + 1 = 2^b \implies 5n + 2 = 2^b
     \]
     Since \(5n + 2\) is a power of 2, let \(5n + 2 = 2^b\). Then:
     \[
     5n = 2^b - 2
     \]
     For \(b = 3\), \(2^b - 2 = 6\), so \(n = 1.2\), which is not an integer. Therefore, there are no solutions in this subcase.

6. Therefore, the only solutions are \(N = 200\) and \(N = 2000\).

The final answer is \(\boxed{200, 2000}\).