1. Interpret the conditions

Let the 4-digit PIN be $a b c d$.

Given:

• All digits are different.
• The greatest digit is $7$.
• The sum of the first two digits equals the sum of the last two digits: $a+b=c+d$

Since the greatest digit is $7$, all digits must be chosen from $\{0,1,2,3,4,5,6,7\}$ and at least one digit must be $7$.

Because all digits are different and $7$ is the greatest digit, this means exactly one of the digits is $7$.

We must count the number of 4-digit arrangements satisfying all conditions.

2. Use the condition $a+b=c+d$

We need 4 distinct digits from $\{0,1,2,3,4,5,6,7\}$ such that they can be split into two pairs with equal sum.

Since one of them is $7$, suppose $7$ is paired with some digit $x$. Then for equality of sums, the other two digits must sum to $7+x.$

So we seek distinct digits $x,y,z$ from $\{0,1,2,3,4,5,6\}$ such that $y+z=7+x$ with all four digits $7,x,y,z$ distinct.

3. Choose the set of 4 digits

We count possible sets containing $7$ for which the other three digits allow equal-sum pairing.

Check $x=0,1,2,3,4,5,6$:

• $x=0$: need $y+z=7$. Possible distinct pairs from $\{1,2,3,4,5,6\}$: $(1,6), (2,5), (3,4)$ giving sets $\{0,1,6,7\},\ \{0,2,5,7\},\ \{0,3,4,7\}$

• $x=1$: need $y+z=8$ with digits from $\{0,2,3,4,5,6\}$. Possible pairs: $(2,6),(3,5)$ giving sets $\{1,2,6,7\},\ \{1,3,5,7\}$

• $x=2$: need $y+z=9$ with digits from $\{0,1,3,4,5,6\}$. Possible pairs: $(3,6),(4,5)$ giving sets $\{2,3,6,7\},\ \{2,4,5,7\}$

• $x=3$: need $y+z=10$ with digits from $\{0,1,2,4,5,6\}$. Possible pair: $(4,6)$ giving set $\{3,4,6,7\}$

• $x=4$: need $y+z=11$ with digits from $\{0,1,2,3,5,6\}$. Possible pair: $(5,6)$ giving set $\{4,5,6,7\}$

• $x=5$: need $y+z=12$ from remaining digits $\{0,1,2,3,4,6\}$, impossible with distinct digits.

• $x=6$: need $y+z=13$ from remaining digits $\{0,1,2,3,4,5\}$, impossible.

Thus the valid digit sets are:

So there are 9 valid sets.

4. Arrange each set into a 4-digit PIN

For each valid set, the digits can be grouped into two equal-sum pairs in exactly one way.

Example: for $\{0,1,6,7\}$, the equal-sum pairing is $7+0=1+6.$

Now place one pair in positions $(a,b)$ and the other in positions $(c,d)$.

For a fixed pairing:

• first pair can be arranged in $2!$ ways,
• second pair can be arranged in $2!$ ways,
• the two pairs can be swapped between the first two and last two positions, giving $2$ choices.

Hence total arrangements per set: $2!\times 2!\times 2=2\times 2\times 2=8.$

So total valid PINs: $9\times 8=72.$

5. Maximum number of trials

In the worst case, he may try all possible valid PINs before getting the correct one.

Therefore, the maximum number of trials necessary is $\boxed{72}.$

6. Comparison with stored answer

Stored correct answer: $72$

Our derived answer: $72$

So the answer agrees with the stored answer.