Determine all pairs $ (a,b)$ of positive integers, each integer having two decimal digits, such that $ 100a\plus{}b$ and $ 201a\plus{}b$ are both perfect squares.
1. Consider the given Diophantine system of equations:
   \[
   100a + b = m^2 \quad \text{(1)}
   \]
   \[
   201a + b = n^2 \quad \text{(2)}
   \]
   where \(10 \leq a, b < 100\) and \(32 \leq m < n < 100\).

2. Subtract equation (1) from equation (2):
   \[
   201a + b - (100a + b) = n^2 - m^2
   \]
   \[
   101a = n^2 - m^2
   \]
   \[
   101a = (n - m)(n + m) \quad \text{(3)}
   \]

3. Since 101 is a prime number, the factors of \(101a\) must include 101. Therefore, \(101\) must divide one of the factors \((n - m)\) or \((n + m)\). Given \(0 < n - m < 101\), it follows that \(101\) must divide \(n + m\).

4. Since \(0 < n + m < 2 \times 101\), the only possibility is:
   \[
   n + m = 101
   \]
   Thus, we have:
   \[
   n = 101 - m
   \]

5. Substitute \(n = 101 - m\) into equation (3):
   \[
   101a = (101 - m)^2 - m^2
   \]
   \[
   101a = 101^2 - 2 \cdot 101m
   \]
   \[
   101a = 101(101 - 2m)
   \]
   \[
   a = 101 - 2m
   \]

6. Substitute \(a = 101 - 2m\) into equation (2) to find \(b\):
   \[
   b = n^2 - 201a
   \]
   \[
   b = (101 - m)^2 - 201(101 - 2m)
   \]
   \[
   b = 101^2 - 2 \cdot 101m + m^2 - 201 \cdot 101 + 402m
   \]
   \[
   b = m^2 + 200m - 10100
   \]

7. Since \(10 \leq b < 100\), we solve for \(m\):
   \[
   10 \leq m^2 + 200m - 10100 < 100
   \]

8. Solving the inequality:
   \[
   10 \leq m^2 + 200m - 10100
   \]
   \[
   m^2 + 200m - 10110 \geq 0
   \]
   \[
   m = 42 \quad \text{(since \(m\) must be an integer and within the range \(32 \leq m < 100\))}
   \]

9. Substitute \(m = 42\) back to find \(a\) and \(b\):
   \[
   a = 101 - 2 \cdot 42 = 17
   \]
   \[
   b = 42^2 + 200 \cdot 42 - 10100 = 1764 + 8400 - 10100 = 64
   \]

10. Verify the solution:
    \[
    100a + b = 100 \cdot 17 + 64 = 1700 + 64 = 1764 = 42^2
    \]
    \[
    201a + b = 201 \cdot 17 + 64 = 3417 + 64 = 3481 = 59^2
    \]

Thus, the only solution to the system of equations is \((a, b) = (17, 64)\).

The final answer is \( \boxed{ (a, b) = (17, 64) } \)