Minimum elements present in {0, 1, 2, ..., 225} to guarantee triple which sums to 225 Suppose I have the set: $$A=\{0, 1, 2, ... 224, 225\}$$
I want to find a triple that sums to $225$ (where a triple is a set of 3 unique values from the set). 
No Repetition Version: 
There are many such triples including: $$(0, 1, 224), (0, 2, 223), ... , (74, 75, 76)$$ Note: $(0, 0, 225)$ is not a valid triple as $0$ is repeated.
What is the minimum number of values which must be present in the set $A$ such that we can guarantee that a triple must exist? 
Or put another way, how many elements can I allow an opponent to strategically remove such that I can still guarantee that a triple must exist (without seeing which values were removed)?
Background & Version with Repetition:
This question arose from an assignment to write an $O(n)$ algorithm which, given a list of perhaps $100,000$ non-negative integers, will determine whether there are $3$ integers which sum to $225$ (note: I have long since finished the assignment, but am still wondering about a better solution). Here, repetition is allowed so $(0, 0, 225)$ or $(75, 75, 75)$ are valid triples.
My approach was to create a 'counter' array $C$ of size $226$ and initialize it with $0's$. While iterating through the list, if the value $i$ is encountered where $0\le i \le 225$, then increment $C[i]$. The counter array therefore keeps track of the number of times each relevant value occurs. 
We can stop counting after $1$ occurrence for any given value $j \gt 225/2$, and after $2$ occurrences for any other value, except $75$ which must be counted for up to $3$ occurrences since the triple $(75, 75, 75)$ must be checked for.
After iterating through the list the counter array can be checked for the presence of valid triples using nested loops and so on. But an optimization can be made. If for example, every value from $0...225$ has been found at least once, then we can say that a triple exists without even checking.
The question is then similar to the simplified case above. After how many 'hits', where some $A[i]$ is incremented, can we simply return true and know that a triple must exist?

I've been trying to find a solution to the simpler version of this problem for quite a while. I asked my algorithms professor and a TA. I also wrote a small program to try and brute force a ceiling on this but quickly ran into combinatorial problems.
 A: $$A=\{0,1,2,\cdots224,225\}$$
Consider this $151$-element subset of $A$:$$B=\{75,76,77,\cdots224,225\}$$
The least-sum triple from this subset is $(75,76,77)$ with a sum of $228$. Adding another element from $\{0,1,2,\cdots73,74\}$ will provide a triple with a sum of $225$.
The minimum number of values which must be present in the set $A$ such that we can guarantee that a triple with sum $225$ must exist is $152$.
A: Logophobic's answer gives the correct answer of $152$, but is missing a complete proof of optimality. Here is a complete proof.
We may partition $A$ into $A = A_1 \cup A_2$, where $A_1 = \{0,1,\ldots,74\}$ and $A_2 = \{75,76,\ldots,225\}$.
Suppose $B \subseteq A$ contains no triple summing to $225$. Let $B_1 = B \cap A_1$ and $B_2 = B \cap A_2$. Let $k$ be the size of $B_1$, so that the elements of $B_1$ are $0\leq x_1 < \cdots < x_k \leq 74$. 
The $k=0,1$ cases were properly analyzed by logophobic, but I will repeat the analysis here for completeness. 
If $k=0$, then $B = B_2 \subseteq A_2$, implying that $|B| \leq |A_2| = 151$. Indeed, the bound is tight, since no 3 distinct elements of $A_2$ sum to $225$. 
If $k=1$, then we demand $B_2 \neq A_2$, since the distinct integers $y_1 = 75$ and $y_2 = 150-x_1$ have the property that $y_1, y_2 \in A_2$ and $x_1+y_1+y_2=225$. And so $|B| = |B_1|+|B_2| \leq 1 + 150 = 151$.
If $k=2$, then consider the distinct elements $y_1=75$ and $y_2=225-x_1-x_2$. Clearly, $y_2\notin B_2$, since $x_1+x_2+y_2=225$. If $y_1\notin B_2$, we have that $|B_2| \leq |A_2|-2 = 149$. If $y_1\in B_2$, then the distinct elements $y_3=150-x_1$ and $y_4=150-x_2$ both cannot be in $B_2$, since $x_1+y_1+y_3=x_2+y_1+y_4=225$. So in this case we also have $|B_2| \leq |A_2|-2 = 149$. Thus, we have $|B| = |B_1|+|B_2| \leq 2 + 149 = 151$.
If $k\geq3$, then note that the $(2k-3)$ distinct values given by $(225-x_i-x_j)$ where $1\leq i < j\leq k$ with $|\{i,j\} \cap \{1,k\}|=1$ all lie in the set $A_2$. This implies that $|B_2|$ is bounded by $|A_2| - (2k-3)$, and so we have $|B| = |B_1| + |B_2| \leq k + (151 - (2k-3)) = 154-k\leq 151$.
This completes the proof.
