Let $m=\lfloor\sqrt{2 n}\rfloor$. We first prove that the sequence of squares $a_{0}+a_{1}, a_{1}+a_{2}$, ... is exactly the sequence $(m+1)^{2},(m+2)^{2}, \ldots$. Then we show that the differences $a_{i}-a_{i-1}$ form a sequence of consecutive even numbers and a sequence of consecutive odd numbers. With this, we can then prove the required statement.
Note that $a_{0}+a_{1}$ is the smallest square greater than $2 n$. Therefore, $a_{0}+a_{1}=(m+1)^{2}$. We now prove by induction on $i$ that $a_{i-1}+a_{i}=(m+i)^{2}$. For $i=1$, we have just proven this. Suppose that $a_{j-1}+a_{j}=(m+j)^{2}$. This means that $a_{j-1} \geq \frac{(m+j-1)^{2}}{2}$ (otherwise, $a_{j}$ could have been chosen such that $a_{j-1}+a_{j}=(m+j-1)^{2}$) and thus

$$ENV0$$\begin{aligned} a_{j} & =(m+j)^{2}-a_{j-1} \\ & \leq(m+j)^{2}-\frac{(m+j-1)^{2}}{2} \\ & =\frac{2 m^{2}+4 m j+2 j^{2}-\left(m^{2}+2 m j+j^{2}+1-2 m-2 j\right)}{2} \\ & =\frac{m^{2}+2 m j+j^{2}-1+2 m+2 j}{2} \\ & a_{j-1}+a_{j}=(m+j)^{2}, so a_{j}+a_{j+1}=(m+j+1)^{2}. This completes the induction. Define now b_{i}=a_{i}-a_{i-1} for all i \geq 1. Then we have \begin{aligned} b_{i+2}-b_{i} & =a_{i+2}-a_{i+1}-a_{i}+a_{i-1} \\ & =\left(a_{i+2}+a_{i+1}\right)+\left(a_{i}+a_{i-1}\right)-2\left(a_{i+1}+a_{i}\right) \\ & =(m+i+2)^{2}+(m+i)^{2}-2(m+i+1)^{2} \\ & =(m+i)^{2}+4(m+i)+4+(m+i)^{2}-2(m+i)^{2}-4(m+i)-2 \\ & =2 \end{aligned}

for all $i \geq 1$. We now see that

$$\left(b_{1}, b_{3}, b_{5}, \ldots\right)=\left(b_{1}, b_{1}+2, b_{1}+4, \ldots\right), \quad\left(b_{2}, b_{4}, b_{6}, \ldots\right)=\left(b_{2}, b_{2}+2, b_{2}+4, \ldots\right)$$

We have $b_{1}+b_{2}=\left(a_{2}-a_{1}\right)+\left(a_{1}-a_{0}\right)=\left(a_{2}+a_{1}\right)-\left(a_{1}+a_{0}\right)=(m+2)^{2}-(m+1)^{2}=2 m+3$. In particular, $b_{1}$ and $b_{2}$ have different parities. Now, at least all numbers with the same parity as $b_{1}$ and at least as large as $b_{1}$ can be written as $a_{k}-a_{k-1}$ for some $k$. The same holds for the numbers with the same parity as $b_{2}$ and at least as large as $b_{2}$. All numbers of the form $a_{k}-a_{\ell}$ with $k \geq \ell+2$ are at least as large as $b_{k}+b_{k-1} \geq b_{1}+b_{2}$ and thus larger than $b_{1}$ and larger than $b_{2}$. In this way, we can no longer generate new numbers. The numbers that cannot be written as $a_{k}-a_{\ell}$ with $k>\ell \geq 0$ are thus exactly the numbers with the same parity as $b_{1}$ and smaller than $b_{1}$ and the numbers with the same parity as $b_{2}$ and smaller than $b_{2}$. This is a total of

$$\left\lfloor\frac{b_{1}-1}{2}\right\rfloor+\left\lfloor\frac{b_{2}-1}{2}\right\rfloor$$

numbers, where within the brackets of the floor function, exactly one of the two times is an integer. We can thus write this as

$$\frac{b_{1}-1}{2}+\frac{b_{2}-1}{2}-\frac{1}{2}=\frac{b_{1}+b_{2}-3}{2}=\frac{2 m}{2}=m \text {. }$$

There are thus exactly $m=\lfloor\sqrt{2 n}\rfloor$ positive integers that cannot be written in the required form.