Let $a,b$ be positive integers. Find, with proof, the maximum possible value of $a\lceil b\lambda \rceil - b \lfloor a \lambda \rfloor$ for irrational $\lambda$.
1. **Claim**: The maximum possible value of \( a \lceil b\lambda \rceil - b \lfloor a\lambda \rfloor \) for irrational \(\lambda\) is \( a + b - \gcd(a, b) \).

2. **Step 1**: Show that no greater value can be achieved.
   - Note that \( a \lceil b\lambda \rceil - b \lfloor a\lambda \rfloor \) must be an integer because the floor and ceiling functions return integers.
   - Since \(\gcd(a, b)\) divides both \(a\) and \(b\), the value must be divisible by \(\gcd(a, b)\).
   - Since \(\lambda\) is irrational, \(b\lambda\) and \(a\lambda\) are not integers. This implies \(\lceil b\lambda \rceil = \lfloor b\lambda \rfloor + 1\) and similarly for \(a\lambda\).

3. **Step 2**: Use the fractional part notation.
   - Define \(\{x\}\) as the fractional part of \(x\), so \(\{x\} = x - \lfloor x \rfloor\).
   - Rearrange the definitions: \(\lfloor x \rfloor = x - \{x\}\) and \(\lceil x \rceil = x + 1 - \{x\}\) if \(x\) is not an integer.
   - Substitute these into the expression:
     \[
     a \lceil b\lambda \rceil - b \lfloor a\lambda \rfloor = a(b\lambda + 1 - \{b\lambda\}) - b(a\lambda - \{a\lambda\})
     \]
     \[
     = ab\lambda + a - a\{b\lambda\} - ab\lambda + b\{a\lambda\}
     \]
     \[
     = a - a\{b\lambda\} + b\{a\lambda\}
     \]

4. **Step 3**: Maximize the expression.
   - To maximize \(a - a\{b\lambda\} + b\{a\lambda\}\), minimize \(a\{b\lambda\}\) and maximize \(b\{a\lambda\}\).
   - The range of \(f(x) = \{x\}\) is \([0, 1)\). Thus:
     \[
     a - a\{b\lambda\} + b\{a\lambda\} < a - a \cdot 0 + b \cdot 1 = a + b
     \]
   - Therefore, \(a - a\{b\lambda\} + b\{a\lambda\}\) cannot be equal to \(a + b\). The largest multiple of \(\gcd(a, b)\) below \(a + b\) is \(a + b - \gcd(a, b)\).

5. **Step 4**: Show that the value \(a + b - \gcd(a, b)\) is achievable.
   - Consider \(c = \frac{a}{\gcd(a, b)}\) and \(d = \frac{b}{\gcd(a, b)}\). This is equivalent to dividing \(a \lceil b\lambda \rceil - b \lfloor a\lambda \rfloor\) by \(\gcd(a, b)\).
   - We need to show:
     \[
     c \lceil d\lambda \rceil - d \lfloor c\lambda \rfloor = c + d - 1
     \]
     for some irrational \(\lambda\).

6. **Step 5**: Repeat the steps with \(c\) and \(d\).
   - We have:
     \[
     c \lceil d\lambda \rceil - d \lfloor c\lambda \rfloor = c - c\{d\lambda\} + d\{c\lambda\}
     \]
     \[
     c - c\{d\lambda\} + d\{c\lambda\} = c + d - 1 \implies d\{c\lambda\} - c\{d\lambda\} = d - 1
     \]
   - For this to be true, \(\{c\lambda\} \ge \frac{d-1}{d}\) and \(c\{d\lambda\} < 1\).

7. **Step 6**: Find \(\lambda\) that satisfies the inequalities.
   - Consider the intervals:
     \[
     \frac{d-1}{d} \le \{c\lambda\} < 1 \implies c\lambda \in \left[\frac{d-1}{d}, 1\right) \cup \left[\frac{2d-1}{d}, 2\right) \cup \cdots
     \]
     \[
     0 \le \{d\lambda\} < \frac{1}{c} \implies d\lambda \in \left[0, \frac{1}{c}\right) \cup \left[1, \frac{c+1}{c}\right) \cup \cdots
     \]
   - Multiply both results by \(cd\):
     \[
     cd\lambda \in \left(d-1, d\right] \cup \left(2d-1, 2d\right] \cup \cdots
     \]
     \[
     cd\lambda \in \left[0, 1\right) \cup \left[c, c+1\right) \cup \cdots
     \]

8. **Step 7**: Find a common interval.
   - Find values \(i\) and \(j\) such that \(ic + 1 = jd\). By Bezout's identity, there exist \(i\) and \(j\) such that \(jd - ic = 1\).
   - Choose \(\lambda\) such that \(cd\lambda = jd - \epsilon\) where \(\epsilon\) is very small.

Thus, we have proven that it is possible to choose a value of \(\lambda\) such that \(a \lceil b\lambda \rceil - b \lfloor a\lambda \rfloor = a + b - \gcd(a, b)\). Since we proved this is the maximum possible value, we are done. 

\(\blacksquare\)

The final answer is \( \boxed{ a + b - \gcd(a, b) } \)