Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0=37$, $a_1=72$, $a_m=0$, and \[a_{k+1}=a_{k-1}-\frac{3}{a_k}\] for $k=1,2, \dots, m-1$. Find $m$.
1. Given the sequence \(a_0, a_1, \ldots, a_m\) with \(a_0 = 37\), \(a_1 = 72\), and \(a_m = 0\), and the recursive relation:
   \[
   a_{k+1} = a_{k-1} - \frac{3}{a_k}
   \]
   for \(k = 1, 2, \ldots, m-1\).

2. We start by examining the condition \(a_m = 0\). From the recursive relation, we have:
   \[
   a_m = a_{m-2} - \frac{3}{a_{m-1}} = 0
   \]
   This implies:
   \[
   a_{m-2} = \frac{3}{a_{m-1}}
   \]
   Multiplying both sides by \(a_{m-1}\), we get:
   \[
   a_{m-2} \cdot a_{m-1} = 3
   \]

3. Next, we use the recursive relation to find a general pattern. For \(k = m-2\), we have:
   \[
   a_{m-1} = a_{m-3} - \frac{3}{a_{m-2}}
   \]
   Substituting \(a_{m-2} = \frac{3}{a_{m-1}}\) into the equation, we get:
   \[
   a_{m-1} = a_{m-3} - \frac{3}{\frac{3}{a_{m-1}}} = a_{m-3} - a_{m-1}
   \]
   Rearranging, we find:
   \[
   2a_{m-1} = a_{m-3}
   \]
   Therefore:
   \[
   a_{m-3} = 2a_{m-1}
   \]

4. Continuing this pattern, we find:
   \[
   a_{m-3} \cdot a_{m-4} = 6
   \]
   \[
   a_{m-4} \cdot a_{m-5} = 9
   \]
   By induction, we can generalize:
   \[
   a_{m-n} \cdot a_{m-n-1} = 3n
   \]

5. Given \(a_0 = 37\) and \(a_1 = 72\), we calculate:
   \[
   a_0 \cdot a_1 = 37 \cdot 72 = 2664
   \]
   According to our pattern:
   \[
   a_{m-888} \cdot a_{m-889} = 3 \cdot 888 = 2664
   \]

6. Since \(a_0 \cdot a_1 = 2664\) and \(a_{m-888} \cdot a_{m-889} = 2664\), we conclude:
   \[
   m - 888 = 1 \implies m = 889
   \]

The final answer is \(\boxed{889}\).