Find all pairs of positive integers \((a,b)\) with the following property: there exists an integer \(N\) such that for any integers \(m\ge N\) and \(n\ge N\), every \(m\times n\) grid of unit squares may be partitioned  into \(a\times b\) rectangles and fewer than \(ab\) unit squares.

[i]Proposed by Holden Mui[/i]
To find all pairs of positive integers \((a, b)\) such that there exists an integer \(N\) with the property that for any integers \(m \ge N\) and \(n \ge N\), every \(m \times n\) grid of unit squares can be partitioned into \(a \times b\) rectangles and fewer than \(ab\) unit squares, we need to analyze the possible values of \(a\) and \(b\).

1. **Basic Cases:**
   - For \((a, b) = (1, 1)\), any \(m \times n\) grid can trivially be partitioned into \(1 \times 1\) rectangles.
   - For \((a, b) = (1, 2)\) or \((2, 1)\), any \(m \times n\) grid can be partitioned into \(1 \times 2\) or \(2 \times 1\) rectangles, respectively, with at most one unit square left over if \(m\) or \(n\) is odd.

2. **Inductive Construction for \(1 \times 3\) Rectangles:**
   - For \((a, b) = (1, 3)\) or \((3, 1)\), we can construct a tiling for \((m+3) \times (n+3)\) from a tiling for \((m, n)\). This can be done by adding a strip of \(1 \times 3\) rectangles along the edge of the existing tiling.

3. **Inductive Construction for \(2 \times 3\) Rectangles:**
   - For \((a, b) = (2, 3)\) or \((3, 2)\), we can construct a tiling for \((m+6) \times (n+6)\) from a tiling for \((m, n)\). This can be done by adding a strip of \(2 \times 3\) rectangles along the edge of the existing tiling.

4. **Completeness of Solution Set:**
   - To show that no other pairs \((a, b)\) work, consider the following claim:
     - **Claim:** For \(c, d \in \mathbb{Z}^+\) and \(k \ge 4\), any tiling of a \((ck+2) \times (dk + \lceil \frac{k}{2} \rceil)\) rectangle with \(1 \times k\) rectangles must leave at least \(k\) cells empty.
     - **Proof:** Consider a \((ck+2) \times (dk + \lceil \frac{k}{2} \rceil)\) grid with its bottom-left cell at the origin, and label each cell with the sum of its coordinates modulo \(k\). Every \(1 \times k\) rectangle must cover one cell whose label is \(k-1\). However, when \(k \ge 4\), no cell in the top-right-most \(2 \times \lceil \frac{k}{2} \rceil\) rectangle can contain a \(k-1\), since the top-right cell's label is \((2-1) + (\lceil \frac{k}{2} \rceil - 1) = \lceil \frac{k}{2} \rceil < k-1\). Therefore, any tiling of \(1 \times k\) rectangles must leave at least \(2 \times \lceil \frac{k}{2} \rceil \ge k\) cells empty. \(\blacksquare\)

5. **Verification for Larger Rectangles:**
   - For \((a, b)\) with \(\max(a, b) \ge 4\), assume \(b \ge 4\). Since \(a \times b\) rectangles can be partitioned into \(a\) separate \(1 \times b\) rectangles, it suffices to find values for \(c\) and \(d\) for which:
     \[
     a \mid \left\lfloor \frac{(bc+2)(bd + \lceil \frac{b}{2} \rceil)}{b} \right\rfloor = bcd + \lceil \frac{b}{2} \rceil c + 2d + 1
     \]
     - If \(b\) is even, then \(a\) must be odd or else large odd-sized grids cannot be partitioned. Choosing \(d = \frac{a-1}{2}\) works.
     - If \(b\) is odd, choosing \(c = k(2d+1)\) gives:
       \[
       bcd + \lceil \frac{b}{2} \rceil c + 2d + 1 = (2d+1)\left(b\left(\frac{c}{2}\right)+1\right)+\frac{c}{2} = (2d+1)\frac{bk(2d+1)+3}{2}
       \]
       Then \(d\) can be chosen such that the first term contains all odd prime factors of \(a\) and \(k\) can be chosen so that the second term contains all even prime factors of \(a\).

6. **Verification for Specific Pairs:**
   - Check that \((2, 2)\) and \((3, 3)\) do not work. \((2, 2)\) does not work because each row in a partitioning of an odd-sized grid must contain a unit square, and \((3, 3)\) does not work for similar reasons.

The final answer is \(\boxed{\{(1, 1), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}}\).