Let $\mathbb N$ be the set of positive integers. Determine all functions $f:\mathbb N\times\mathbb N\to\mathbb N$ that satisfy both of the following conditions:
[list]
[*]$f(\gcd (a,b),c) = \gcd (a,f(c,b))$ for all $a,b,c \in \mathbb{N}$.
[*]$f(a,a) \geq a$ for all $a \in \mathbb{N}$.
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1. **Define the problem and notation:**
   We are given a function \( f: \mathbb{N} \times \mathbb{N} \to \mathbb{N} \) that satisfies:
   \[
   f(\gcd(a, b), c) = \gcd(a, f(c, b)) \quad \text{for all } a, b, c \in \mathbb{N}
   \]
   and
   \[
   f(a, a) \geq a \quad \text{for all } a \in \mathbb{N}.
   \]

2. **Initial substitution:**
   Let \( P(a, b, c) \) denote the assertion of the first condition:
   \[
   P(a, b, c): f(\gcd(a, b), c) = \gcd(a, f(c, b)).
   \]

3. **Evaluate \( P(a, a, a) \):**
   \[
   P(a, a, a): f(a, a) = \gcd(a, f(a, a)).
   \]
   Since \( \gcd(a, f(a, a)) \leq a \) and \( f(a, a) \geq a \) (from the second condition), we must have:
   \[
   f(a, a) = a.
   \]

4. **Evaluate \( P(a, ab, ab) \):**
   \[
   P(a, ab, ab): f(a, ab) = \gcd(a, f(ab, ab)).
   \]
   Since \( f(ab, ab) = ab \) (from the previous step), we get:
   \[
   f(a, ab) = \gcd(a, ab) = a.
   \]

5. **General case:**
   Consider \( P(a, ck, c) \):
   \[
   f(\gcd(a, ck), c) = \gcd(a, f(c, k)).
   \]
   Define \( d = \gcd(a, k, c) \), \( \gcd(a, c) = dx \), and \( \gcd(a, k) = dy \). Let:
   \[
   a = dxy a_1, \quad c = dxc_1, \quad k = dyk_1,
   \]
   where \( \gcd(a_1, x) = \gcd(a_1, y) = 1 \) and \( \gcd(y, c_1) = 1 \).

6. **Simplify the relation:**
   Using the above definitions, we have:
   \[
   f(\gcd(a, ck), c) = f(dxy, dxc_1) = dx.
   \]
   This implies:
   \[
   f(a, b) = \gcd(a, b).
   \]

7. **Verification:**
   We need to verify that \( f(a, b) = \gcd(a, b) \) satisfies both conditions:
   - For the first condition:
     \[
     f(\gcd(a, b), c) = \gcd(\gcd(a, b), c) = \gcd(a, \gcd(b, c)) = \gcd(a, f(c, b)).
     \]
   - For the second condition:
     \[
     f(a, a) = \gcd(a, a) = a \geq a.
     \]

Thus, the function \( f(a, b) = \gcd(a, b) \) satisfies both conditions.

The final answer is \( \boxed{ f(a, b) = \gcd(a, b) } \).