Ms. Janet, you made a mistake!

Let the square number be $n^2$ .When $n\le50$ ,the difference of two adjacent square numbers is not larger than 100,so the integers from 10 to 25 are all possible first two digits.

When $n>50$ ,the difference of two adjacent square numbers is larger than 100,so no two square numbers larger than 2500 have the same first two digits.

There are only 50 square numbers from 2500 to 9999( $n=51 \text{ to }99$ ),but from 2500 to 9999,there are 75 cases of the first two digits(25,26,27...98,99).Hence,there are $75-50=25$ impossible first two digits of a four digit square number.

Relevant wiki: Modular Arithmetic

$\textbf{Note}$ My solution is pretty long. If you don't feel like going through mine, you can just read @XX solution as his is short, sweet and complete, which I give full credits to.

Jerry knew Ms. Janet made a mistake as soon as she wrote down the second digit because the square of a positive integer which goes up to four-digit could not yield the concatenation of the first two digits written down.

For example, $59^2=3481,60^2=3600$ , we see that $35$ is "skipped" so $35$ may be the concatenation of the first two digits written down.

But it would be pretty tedious if we compute the square of all positive two-digit numbers(since $100^2=10000$ , which is the smallest five-digit positive integers) and see which two-digit numbers are skipped. So, is there a better way than that? The answer is yes and no computer programming is needed. (Although you can solve this with coding)

But before that, we will have to prove some useful statements.

$\textbf{Statement 1:}$ If $(50+x)^2\equiv p(mod100), then x^2\equiv p(mod100)$ for $x\in Z$

$\textbf{Proof:}$

$(50+x)^2\equiv p(mod100)$

$2500+100x+x^2\equiv p(mod100)$

$x^2\equiv p-2500-100x(mod100)$

$x^2\equiv p(mod100)$

This completes the proof.

The converse of the statement is also true.

$\textbf{Statement 2:}$ If $x^2\equiv p(mod100)$ , then $(50-x)^2\equiv p(mod100)$ for $x\in Z$

The proof is similiar to the first one, so I will skip the proof.

Now, we shall go deeper into reasoning of why some of the two-digit numbers are skipped. For example, we see that $59^2=3481,60^2=3600$ , $35$ is skipped because $59^2\equiv81(mod100)$ , $59^2-81=3400$ , $60^2=3600$ , $3600-3400=200\geq200$ and so the $35$ is skipped.

Now, we shall move on to the generalization.

If $x^2\equiv p(mod100)$ and $(x+1)^2-(x^2-p)\geq200$ , then there will be skipped two-digit numbers.

$\Rightarrow$ If $2x+1+p\geq 200$ , then there will be a skipped two-digit numbers.

Notice that $x\geq 50$ because $p\leq99$ and if $x<50\Rightarrow x\leq49$ , $2x+1\leq99$

$\Rightarrow 2x+1+p\leq198<200$

However, we must also show that between each perfect squares, there will be at most $1$ skipped two-digit number.

$\textbf{Proof:}$ $x\leq99,p\leq99$ , $2x+1+p\leq 2(99)+1+99=298<300$ , this completes the proof.

$x\geq 50$ , we can let $x=50+y$

$\Rightarrow 2(50+y)+1+p\geq200$

$100+2y+1+p\geq200$

$2y+1+p\geq100$

$50\leq x\leq99 \Rightarrow 50\leq 50+y\leq99 \Rightarrow 0\leq y\leq49$

$(50+y)^2\equiv p(mod100) \Rightarrow y^2\equiv p(mod100)$ ( $\textbf{Statement 1}$ )

We will divide the numbers to be verified into several cases.

Generally speaking, if $p\geq 50$ , the value of $y$ is probably the valid solution we are looking for. However , there are some counter-examples too. For instance, when $y=13, p=69$ , $2y+1+p=2(13)+1+69=96<100$ and when $y=29, p=41$ , $2y+1+p=2(29)+1+41=100\geq100$ . In the later cases, you can observe that even when $p<50$ , many of their corresponding values of $y$ are valid solutions. This is so because $y$ gets progressively larger to help fill the gap left by smaller $p$ . Also, the case where $p\geq 50$ but the corresponding value of $y$ is not valid solution will diminish in the later cases.

$\textbf{Case1:}$ If $0\leq y\leq10$ ,

After some verification,

we conclude that $y=9$ is the solution in this case.

When $y=9$ , $2y+1+p=2(9)+1+81=100\geq100$

$\textbf{Case2:}$ If $11\leq y\leq20$ ,

After some verification,

we conclude that $y=14,17,19$ are the solutions in this case.

When $y=14$ , $2y+1+p=2(14)+1+96\geq100$

When $y=17$ , $2y+1+p=2(17)+1+89\geq100$

When $y=19$ , $2y+1+p=2(19)+1+61\geq100$

$\textbf{Case3:}$ If $21\leq y\leq30$ ,

After some verification,

we conclude that $y=22,24,26,28,29$ are the solutions in this case.

When $y=22$ , $2y+1+p=2(22)+1+84\geq100$

When $y=24$ , $2y+1+p=2(24)+1+76\geq100$

When $y=26$ , $2y+1+p=2(26)+1+76\geq100$

When $y=28$ , $2y+1+p=2(28)+1+84\geq100$

When $y=29$ , $2y+1+p=2(29)+1+41\geq100$

$\textbf{Case3:}$ If $31\leq y\leq40$ ,

Here, we can utilize $\textbf{Statement 2}$ to further simplify our computations.

For example, when $y=34, y^2=34^2$ . $16^2\equiv56(mod100)$ . $\textbf{Statement 2}$ tells us that $34^2=(50-16)^2\equiv56(mod100)$

After some verification,

we conclude that $y=31,32,34,36,37,38,39$ are the solutions in this case.

When $y=31$ , $2y+1+p=2(31)+1+61\geq100$

When $y=33$ , $2y+1+p=2(33)+1+89\geq100$

When $y=34$ , $2y+1+p=2(34)+1+56\geq100$

When $y=36$ , $2y+1+p=2(36)+1+96\geq100$

When $y=37$ , $2y+1+p=2(37)+1+69\geq100$

When $y=38$ , $2y+1+p=2(38)+1+44\geq100$

When $y=39$ , $2y+1+p=2(39)+1+21\geq100$

$\textbf{Case3:}$ If $31\leq y\leq40$ ,

Similarly, here, we can utilize $\textbf{Statement 2}$ to further simplify our computations.

After some verification,

we conclude that $y=41,42,43,44,45,46,47,48,49$ are the solutions in this case.

When $y=41$ , $2y+1+p=2(41)+1+81\geq100$

When $y=42$ , $2y+1+p=2(42)+1+64\geq100$

When $y=43$ , $2y+1+p=2(43)+1+49\geq100$

When $y=44$ , $2y+1+p=2(44)+1+36\geq100$

When $y=45$ , $2y+1+p=2(45)+1+25\geq100$

When $y=46$ , $2y+1+p=2(46)+1+16\geq100$

When $y=47$ , $2y+1+p=2(47)+1+9\geq100$

When $y=48$ , $2y+1+p=2(48)+1+4\geq100$

When $y=49$ , $2y+1+p=2(49)+1+1\geq100$

So, the number of possible solutions $=1+3+5+7+9=\boxed{25}$

Whew! That was a pretty long solution.