Four. (20 points) Let $n$ be a positive integer greater than 1. If there exists a positive integer $i (i < n)$ such that
$$
\sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{n} b^{2}
$$

is a perfect square, then $n$ is called a "TI number." For example, by $\left(1^{2}+2^{2}+\cdots+9^{2}\right)-\left(10^{2}+11^{2}\right)=8^{2}$, we know that 11 is a TI number. Question: Are 46, 36, and 26 TI numbers? Why?
(1) From $\sum_{a=1}^{39} a^{2}-\sum_{b=40}^{46} b^{2}=7569=87^{2}$, we know that 46 is a TI number.
(2) From $\sum_{a=1}^{30} a^{2}-\sum_{b=31}^{36} b^{2}=52^{2}$, we know that 36 is a TI number.
(3) By calculation, when $1 \leqslant i \leqslant 20$,
$$
\sum_{a=1}^{i} a^{2}-\sum_{b=i+1}^{26} b^{2}<0 \text {, }
$$

and $\square$
$$
\begin{array}{l}
\sum_{a=1}^{21} a^{2}-\sum_{b=2}^{2 \infty} b^{2}=421, \\
\sum_{a=1}^{2} a^{2}-\sum_{b=3}^{\infty} b^{2}=1389, \\
\sum_{a=1}^{2} a^{2}-\sum_{b=24}^{\infty} b^{2}=2447, \\
\sum_{a=1}^{24} a^{2}-\left(25^{2}+26^{2}\right)=3599, \\
\sum_{a=1}^{25} a^{2}-26^{2}=4849,
\end{array}
$$

none of them are perfect squares.
Therefore, 26 is not a TI number.