At a party, there are $100$ cats. Each pair of cats flips a coin, and they shake paws if and only if the coin comes up heads. It is known that exactly $4900$ pairs of cats shook paws. After the party, each cat is independently assigned a ``happiness index" uniformly at random in the interval $[0,1]$. We say a cat is [i]practical[/i] if it has a happiness index that is strictly greater than the index of every cat with which it shook paws. The expected value of the number of practical cats is $\frac{m}{n}$, where $m$ and $n$ are positive integers with $\gcd(m,n)=1$. Compute $100m + n$.

[i]Proposed by Brandon Wang[/i]
1. **Understanding the Problem:**
   - There are 100 cats at a party.
   - Each pair of cats flips a coin and shakes paws if the coin comes up heads.
   - There are exactly 4900 pairs of cats that shook paws.
   - Each cat is assigned a happiness index uniformly at random in the interval \([0,1]\).
   - A cat is considered practical if its happiness index is strictly greater than the index of every cat with which it shook paws.
   - We need to find the expected number of practical cats and express it in the form \(\frac{m}{n}\) where \(\gcd(m,n)=1\), and then compute \(100m + n\).

2. **Setting Up the Problem:**
   - Let \(X_i\) be an indicator random variable that is 1 if cat \(i\) is practical and 0 otherwise.
   - The expected number of practical cats is \(E\left[\sum_{i=1}^{100} X_i\right] = \sum_{i=1}^{100} E[X_i]\) by the linearity of expectation.

3. **Calculating \(E[X_i]\):**
   - For cat \(i\) to be practical, its happiness index must be greater than the happiness index of every cat it shook paws with.
   - Let \(d_i\) be the degree of cat \(i\) in the graph where an edge represents a pair of cats that shook paws.
   - The probability that cat \(i\) is practical is the probability that its happiness index is the maximum among \(d_i + 1\) values (its own index and the indices of the \(d_i\) cats it shook paws with).
   - Since the happiness indices are uniformly distributed, the probability that cat \(i\) has the highest index among \(d_i + 1\) values is \(\frac{1}{d_i + 1}\).

4. **Expected Degree Calculation:**
   - The total number of edges in the graph is 4900, and there are 100 cats.
   - The sum of the degrees of all cats is \(2 \times 4900 = 9800\) (since each edge is counted twice).
   - The average degree \(\bar{d}\) of a cat is \(\frac{9800}{100} = 98\).

5. **Expected Value of Practical Cats:**
   - The expected value of \(d_i\) for a randomly chosen cat is 98.
   - Therefore, the expected value of \(\frac{1}{d_i + 1}\) is approximately \(\frac{1}{98 + 1} = \frac{1}{99}\).
   - The expected number of practical cats is \(100 \times \frac{1}{99} = \frac{100}{99}\).

6. **Simplifying the Fraction:**
   - The fraction \(\frac{100}{99}\) is already in its simplest form since \(\gcd(100, 99) = 1\).
   - Thus, \(m = 100\) and \(n = 99\).

7. **Final Calculation:**
   - Compute \(100m + n = 100 \times 100 + 99 = 10000 + 99 = 10099\).

The final answer is \(\boxed{10099}\).