Let $A$ be the set of the $16$ first positive integers. Find the least positive integer $k$ satisfying the condition: In every $k$-subset of $A$, there exist two distinct $a, b \in A$ such that $a^2 + b^2$ is prime.
1. **Define the set \( A \) and the problem constraints:**
   Let \( A = \{1, 2, 3, \ldots, 16\} \). We need to find the smallest integer \( k \) such that in every \( k \)-subset of \( A \), there exist two distinct elements \( a \) and \( b \) in \( A \) such that \( a^2 + b^2 \) is prime.

2. **Check if \( k = 8 \) satisfies the condition:**
   Consider the subset \( B = \{2, 4, 6, 8, 10, 12, 14, 16\} \). All elements in \( B \) are even. For any two distinct elements \( a, b \in B \), \( a^2 + b^2 \) is even and divisible by 4. Since the only even prime number is 2, and \( a^2 + b^2 \geq 4 \) for any \( a, b \in B \), \( a^2 + b^2 \) cannot be prime. Therefore, \( k = 8 \) does not satisfy the condition.

3. **Check if \( k = 9 \) satisfies the condition:**
   Consider the following pairs of elements from \( A \):
   \[
   (15, 2), (11, 4), (5, 6), (13, 8), (3, 10), (7, 12), (9, 14), (1, 16)
   \]
   We need to verify that \( a^2 + b^2 \) is prime for each pair:
   \[
   \begin{align*}
   15^2 + 2^2 &= 225 + 4 = 229 \quad (\text{prime}) \\
   11^2 + 4^2 &= 121 + 16 = 137 \quad (\text{prime}) \\
   5^2 + 6^2 &= 25 + 36 = 61 \quad (\text{prime}) \\
   13^2 + 8^2 &= 169 + 64 = 233 \quad (\text{prime}) \\
   3^2 + 10^2 &= 9 + 100 = 109 \quad (\text{prime}) \\
   7^2 + 12^2 &= 49 + 144 = 193 \quad (\text{prime}) \\
   9^2 + 14^2 &= 81 + 196 = 277 \quad (\text{prime}) \\
   1^2 + 16^2 &= 1 + 256 = 257 \quad (\text{prime})
   \end{align*}
   \]
   Since each pair sums to a prime number, any \( 9 \)-subset of \( A \) must include at least one of these pairs. Therefore, \( k = 9 \) satisfies the condition.

The final answer is \( \boxed{ k = 9 } \).