We are given five equal-looking weights of pairwise distinct masses.  For any three weights $A$, $B$, $C$, we can check by a measuring if $m(A) < m(B) < m(C)$, where $m(X)$ denotes the mass of a weight $X$ (the answer is [i]yes[/i] or [i]no[/i].)  Can we always arrange the masses of the weights in the increasing order with at most nine measurings?
1. **Label the weights**: Let the weights be labeled as \( A, B, C, D, E \).

2. **Determine the order of three weights**: We need to determine the order of \( A, B, \) and \( C \). There are \( 3! = 6 \) possible permutations of these three weights. We can determine the correct order using at most 5 comparisons:
   - Compare \( A, B, C \) to see if \( m(A) < m(B) < m(C) \).
   - If not, compare \( A, C, B \) to see if \( m(A) < m(C) < m(B) \).
   - Continue this process until the correct order is found.

3. **Assume the order of \( A, B, C \) is \( P, Q, R \) such that \( m(P) < m(Q) < m(R) \)**.

4. **Determine the position of \( D \)**:
   - Compare \( P, D, R \):
     - If \( m(P) < m(D) < m(R) \), then compare \( Q, D, R \):
       - If \( m(Q) < m(D) < m(R) \), the order is \( P, Q, D, R \).
       - If not, the order is \( P, D, Q, R \).
     - If not, compare \( D, P, R \):
       - If \( m(D) < m(P) < m(R) \), the order is \( D, P, Q, R \).
       - If not, the order is \( P, Q, R, D \).

5. **Assume the order of \( A, B, C, D \) is \( W, X, Y, Z \) such that \( m(W) < m(X) < m(Y) < m(Z) \)**.

6. **Determine the position of \( E \)**:
   - Compare \( W, E, Z \):
     - If \( m(W) < m(E) < m(Z) \), then compare \( X, E, Y \):
       - If \( m(X) < m(E) < m(Y) \), the order is \( W, X, E, Y, Z \).
       - If not, the order is \( W, X, Y, E, Z \).
     - If not, compare \( E, W, Z \):
       - If \( m(E) < m(W) < m(Z) \), the order is \( E, W, X, Y, Z \).
       - If not, the order is \( W, X, Y, Z, E \).

7. **Count the number of comparisons**:
   - Determining the order of \( A, B, C \) takes at most 5 comparisons.
   - Determining the position of \( D \) takes at most 2 comparisons.
   - Determining the position of \( E \) takes at most 2 comparisons.

Thus, the total number of comparisons is \( 5 + 2 + 2 = 9 \).

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