(a) It is clear that it is enough to show that if $n \geq 4$, then $n^{2}$ cannot be written as the sum of $n^{2}-13$ positive squares. Suppose it can be written as:

$$n^{2}=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n^{2}-13}^{2}$$

where $a_{1}, \cdots, a_{n^{2}-13} \geq 1$. Rearrange this as follows:

$$\sum_{i=1}^{n^{2}-13}\left(a_{i}^{2}-1\right)=13$$

If we show that 13 cannot be written as the sum of numbers of the form $x^{2}-1$ (where $x \geq 1$), we are done. Since for $x \geq 4$, $x^{2}-1>13$, the values of $a_{i}$ can only be 1, 2, and 3, meaning that on the left side of (1), every term is 0, 3, or 8.

If no 8 appears, then the sum is divisible by 3; if one 8 appears, then the sum gives a remainder of 2 when divided by 3; if at least two 8s appear, then the sum is at least 16. Since 13 gives a remainder of 1 when divided by 3 and is less than 16, the sum cannot be 13.

However, it is true - and we will use this in part (b) - that all integers greater than 13 can be written as the sum of numbers of the form $x^{2}-1$, and even as the sum of 3s and 8s.

(b) Clearly, we need to find an $n$ such that $n^{2}$ can be written as the sum of more than two positive squares, i.e., $n$ is the largest element of a Pythagorean triple. The smallest such numbers are:

$$5, \quad 10, \quad 13, \quad 15, \quad 17$$

It is easy to check that the squares of 5 and 10 cannot be written as the sum of three squares, so $n=5$ and $n=10$ are not suitable. However, we will show that $n=13$ is suitable. For this, since we have already proven statement (a), it is enough to show that $S(13) \geqslant 13^{2}-14=155$, i.e., $13^{2}=169$ can be written as the sum of 1, 2, 3, ..., 154, or exactly 155 positive squares.

Let $k \leq 155$ and try to write 169 as the sum of $k$ positive squares:

$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}=169$$

where $a_{1}, \ldots, a_{k} \geq 1$. Rearrange this similarly to (1):

$$\sum_{i=1}^{k}\left(a_{i}^{2}-1\right)=169-k$$

Now we need to write the number $169-k$ as the sum of $k$ numbers of the form $x^{2}-1$. The numbers of the form $x^{2}-1$ that are less than 169 are:

$$0,3,8,15,24,35,48,63,80,99,120,143,168$$

Let $a_{1}$ be the largest positive integer such that

$$a_{1}^{2}-1 \leq 155-k$$

(such an integer exists because $155-k \geq 0$), i.e.,

$$$$
a_{1}^{2}-1= \begin{cases}0, & \text { if } 14 \leq 169-k \leq 16 \\ 3, & \text { if } 17 \leq 169-k \leq 21 \\ 8, & \text { if } 22 \leq 169-k \leq 28 \\ 15, & \text { if } 29 \leq 169-k \leq 37 \\ 24, & \text { if } 38 \leq 169-k \leq 48 \\ 35, & \text { if } 49 \leq 169-k \leq 61 \\ 48, & \text { if } 62 \leq 169-k \leq 76