Suppose that we have $N+1$ apples (each at least $25$ grams) with a total weight of $100N$ grams, and we want to cut them into pieces of at least $25$ grams so that we can make $N$ groups weighing $100$ grams each.
If $N=1$, this is trivial: we have two pieces totalling $100$ grams, and we just put them together.
If $N>1$, I will show that we can always reduce the situation to make $N$ smaller.
There are three possibilities:
- The smallest apple and the largest apple together weigh at least $125$ grams, and the smallest apple weighs no more than $75$ grams;
- The smallest apple and the largest apple together weigh less than $125$ grams;
- Every apple weighs more than $75$ grams.
Case 1: The smallest apple and the largest apple together weigh at least $125$ grams, and the smallest apple weighs no more than $75$ grams.
In this case, we can cut a piece off the larger apple so that the piece and the smallest apple weigh $100$ grams together. We have gotten rid of one apple and one set of $100$ grams, so we have reduced $N$ by $1$.
Case 2: The smallest apple and the largest apple together weigh less than $125$ grams.
Because all the apples' weights sum to $100N$, the rest of the apples must have a total weight of at least $100N-125$ grams. The average weight of these apples is at least $\frac{100N-125}{N-1}=100-\frac{25}{N-1}$, which is at least $75$. Therefore the second-largest and largest apples must each weigh at least $75$ grams, and the smallest apple must weigh less than $50$ grams. We can cut a piece off the larger apple so that the piece and the smallest apple weigh $75$ grams together, then cut $25$ grams from the second-largest apple, for a total of $100$ grams. Again, we have gotten rid of one apple and one set of $100$ grams, so we have reduced $N$ by $1$.
Case 3: Every apple weighs more than $75$ grams.
Let the weight of the largest apple be $x$. For each $m\le N$, let $W_m$ be the weight of the $m$-th smallest apple, let $S_m$ be the sum of the weights of the $m$ smallest apples, and let $D_m=100m-S_m$. $D_m$ represents the size of the piece that would need to be added to the first $m$ apples to make them total $100m$ grams. In particular, $D_0=0$ and $D_N=x$.
$D_m-D_{m-1}=100+S_{m-1}-S_m=100-W_m<25$, so $D_m$ cannot increase by more than $25$ grams at a time. Because $x>75$, the difference between $25$ and $x-25$ is more than $25$ grams. Because $D_0<25$ and $D_N>x-25$, some $D_m$ will necessarily have a value between $25$ and $x-25$. Then we can cut off a piece of size $D_m$ from the largest apple and combine it with the smallest $m$ apples to make $m+1$ pieces totalling $100m$ grams. The remaining $N-m+1$ pieces total $100(N-m)$ grams. This reduces our situation to two smaller situations.
Therefore, for all $N$, if we have $N+1$ apples with a total weight of $100N$ grams, we can divide them so that we have $N$ groups of $100$ grams.
We start with $100$ apples. The largest one is at least $100$ grams, so we can safely cut it in half. Then we have $101$ pieces totalling $10000$ grams. This corresponds to $N=100$. As just shown, we can cut these pieces to make $100$ groups of $100$ grams.